Torque on a Current-Loop : Moving-Coil Galvanometer

the plane of the coil remains parallel to the lines of forces in every position

Reason — In a radial magnetic field the lines of force are directed along the radii of the cylindrical soft-iron core. Hence, in every position of the coil its plane remains parallel to the lines of force, so that θ = 90° and the deflecting torque always has its maximum value NIAB.

all of the above

Reason — The current sensitivity of a moving coil galvanometer is

$\frac{\phi}{\text I} = \frac{\text{NAB}}{\text c}$

Hence, the sensitivity increases on increasing the number of turns N, the magnetic field intensity B and the area A of the coil.

all of the above

Reason — Phosphor-bronze is a good conductor of electricity, does not oxidise (is rust proof), has a small torsional constant (torsion couple per unit twist) and has very less elastic after effect. All these properties make it suitable for the suspension strip.

$\frac{\text G}{19}$

Reason — The current through the galvanometer is I_g = 5% of I = $\frac{\text I}{20}$, so the shunt carries the rest, $\frac{19\text I}{20}$. As G and S are in parallel,

$$\begin{aligned}\text I_\text g \times \text G = (\text I - \text I_\text g) \times \text S \\[1em] \frac{\text I}{20} \times \text G = \frac{19\text I}{20} \times \text S \\[1em] \text S = \frac{\text G}{19}\end{aligned}$$

less than galvanometer

Reason — An ammeter is formed by connecting a small resistance S (shunt) in parallel with the galvanometer coil G. The resistance of the ammeter, $\text R_\text A = \frac{\text{GS}}{\text G + \text S}$, is less than even the shunt S, and hence much less than G.

a high resistance in series with its coil

Reason — To convert a galvanometer into a voltmeter its resistance must be increased. So a high resistance $\text R = \frac{\text V}{\text I_\text g} - \text G$ is connected in series with its coil.

zero

Reason — When B is perpendicular to the plane of the loop, the normal to the loop is along B, i.e., θ = 0°. Hence, $\tau = \text{BIA}\sin 0^\circ = 0$.

10^-4 N-m

Reason — The field is parallel to the plane of the loop, so the normal to the loop makes θ = 90° with the field and the torque is maximum.

$$\begin{aligned}\tau = \text{IAB}\sin 90^\circ \\[1em] = 10.0 \times (1.0 \times 10^{-4}) \times 0.1 \times 1 \\[1em] \tau = 10^{-4}\ \text{N m}\end{aligned}$$

The magnitude of magnetic moment now diminishes

Reason — Initially the magnetic moment is m = IπR², along the z-axis. After bending, each half is a semicircle of area $\frac{\pi\text R^2}{2}$. The half in the x-y plane has a moment $\frac{\text m}{2}$ along the z-axis and the half in the y-z plane has a moment $\frac{\text m}{2}$ along the x-axis. The resultant moment is

$$\begin{aligned}\text m' = \sqrt{\left(\frac{\text m}{2}\right)^2 + \left(\frac{\text m}{2}\right)^2} \\[1em] \text m' = \frac{\text m}{\sqrt{2}}\end{aligned}$$

which is less than m. Hence, the magnitude of the magnetic moment diminishes.

zero

Reason — The magnetic moment M is along the axis of the loop, i.e., perpendicular to its plane. Rotating the loop about this axis does not change the direction of M, so the angle θ between M and B remains the same. The work done, $\text W = \text{MB}(\cos\theta_1 - \cos\theta_2)$, is therefore zero.

$\text G(\text n - 1)$

Reason — The current for full-scale deflection is $\text I_\text g = \frac{\text V}{\text G}$. For the range nV with a series resistance R,

$$\begin{aligned}\text I_\text g = \frac{\text{nV}}{\text G + \text R} \\[1em] \frac{\text V}{\text G} = \frac{\text{nV}}{\text G + \text R} \\[1em] \text G + \text R = \text{nG} \\[1em] \text R = \text G(\text n - 1)\end{aligned}$$

Given,

• Current, I = 0.50 A
• Magnetic field, B = 0.20 T
• Radius of the loop, r = 4.9 cm = 4.9 × 10^-2 m

The torque on a current-loop in a uniform magnetic field is τ = IAB sin θ, where θ is the angle between the normal to the plane of the loop and the field. As the plane of the loop is perpendicular to $\vec{\text B}$, the normal is along $\vec{\text B}$, so θ = 0°.

$$\begin{aligned}\tau = \text I \times \pi\text r^2 \times \text B \times \sin 0^\circ \\[1em] = 0.50 \times \pi \times \left(4.9 \times 10^{-2}\right)^2 \times 0.20 \times 0 \\[1em] \tau = 0\end{aligned}$$

Hence, the torque exerted by the magnetic field on the coil is zero.

The radial magnetic field is achieved by using cylindrically-cut (concave) pole-pieces of a strong horse-shoe magnet and placing a soft-iron cylindrical core within the coil, without touching it.

A radial magnetic field is one in which the lines of force are directed along the radii of the cylindrical core, so that the plane of the coil remains parallel to the magnetic field in every position of the coil (θ = 90°). The deflecting torque NIAB on the coil is then independent of its orientation.

The pole-pieces are made cylindrical (concave) so that, together with the soft-iron core, they make the magnetic field radial. The plane of the coil then remains parallel to the field in every position, the deflecting torque is always NIAB and the current is directly proportional to the deflection (I ∝ φ).

Infinite.

Explanation:
An ideal voltmeter draws no current from the circuit, so that the potential difference to be measured is not changed. This requires its resistance to be infinite.

No.

Explanation:
The shunt required to convert a galvanometer into an ammeter of range I is

$\text S = \left(\frac{\text I_\text g}{\text I - \text I_\text g}\right)\text G$

If I < I_g, then (I − I_g) is negative and S would be negative, which is not possible. A shunt can only divert a part of the current, so the range of the ammeter is always greater than I_g.

Given,

• emf of the cell, E = 1.5 V
• Resistance of the voltmeter, R_V = 1500 Ω
• Resistance of the ammeter is negligible

The ammeter of negligible resistance is connected in parallel with R, so it short-circuits R and the whole current passes through the ammeter. The circuit then contains only the voltmeter (1500 Ω) in series with the cell, whose internal resistance is taken as negligible.

By Ohm's law, the current in the circuit is

$$\begin{aligned}\text I = \frac{\text E}{\text R_\text V} \\[1em] = \frac{1.5}{1500} \\[1em] = 1.0 \times 10^{-3}\ \text A \\[1em] \text I = 1.0\ \text{mA}\end{aligned}$$

The voltmeter reads the potential difference across its own resistance,

$$\begin{aligned}\text V = \text I\text R_\text V \\[1em] = 1.0 \times 10^{-3} \times 1500 \\[1em] \text V = 1.5\ \text V\end{aligned}$$

Hence, the ammeter reads 1.0 mA and the voltmeter reads 1.5 V.

A moving coil galvanometer is converted into a voltmeter by connecting a high resistance R in series with its coil. For a voltmeter of range 0 to V volt, $\text R = \frac{\text V}{\text I_\text g} - \text G$, where G is the resistance of the galvanometer and I_g the current for full-scale deflection.

A moving-coil galvanometer works on the principle that when an electric current flows in a coil placed in a magnetic field, a deflecting torque acts upon the coil whose magnitude depends upon the strength of the current (τ = NIAB). In a radial field the deflection of the coil is directly proportional to the current (I ∝ φ), so the current can be measured from the deflection.

There are two reasons :

(i) Aluminium is a light metal. Hence, a large coil can be suspended by a thin strip, as is necessary for sensitivity.

(ii) When the coil moves, eddy currents are developed in the metallic aluminium frame which oppose the motion of the coil. Hence, on stopping the current, the coil comes to rest rapidly.

Phosphor-bronze is used for the suspension fibre because :

(i) its torsional constant is small, as required for the sensitivity of the galvanometer;

(ii) its tensile strength is large enough to support the weight of the coil even when the fibre is very fine;

(iii) it is rust proof;

(iv) it is a good conductor of electricity;

(v) it has very less elastic after effect.

The coil of a moving-coil galvanometer is suspended in a very strong magnetic field produced by the pole-pieces of a powerful horse-shoe magnet and concentrated by the soft-iron core. The earth's magnetic field is negligible in comparison, so it does not affect the working of the galvanometer.

Radial.

Explanation:
The pole-pieces are cylindrically cut and a soft-iron core is placed within the coil, so the lines of force are directed along the radii of the core. The plane of the coil therefore remains parallel to the magnetic field in every position of the coil.

In a radial magnetic field the plane of the coil remains parallel to the lines of force in all orientations of the coil (θ = 90°), so the deflecting torque NIAB remains constant for a given current. In equilibrium NIAB = cφ, i.e., I ∝ φ. Hence, the deflection of the galvanometer is directly proportional to the current in its coil and the scale is uniform.

Soft iron.

Explanation:
The core is a cylindrical piece of soft iron fixed within the coil without touching it. Due to the large permeability of soft iron, it concentrates the lines of force and makes the magnetic field radial and strong.

The soft-iron core :

(i) helps in making the magnetic field radial between the magnetic pole-pieces, so that the plane of the coil is always parallel to the field;

(ii) concentrates the lines of force due to its large permeability, thus making the magnetic field strong and the galvanometer more sensitive.

The current sensitivity of a moving-coil galvanometer is defined as the deflection produced in the galvanometer when a unit current flows through it,

$\frac{\phi}{\text I} = \frac{\text{NAB}}{\text c}$

where N is the number of turns, A the area of the coil, B the magnetic field and c the torsional constant of the suspension. Its unit is radian/ampere.

The current sensitivity of a moving-coil galvanometer is $\frac{\phi}{\text I} = \frac{\text{NAB}}{\text c}$. Two factors on which it depends are :

(i) the magnetic field B — the sensitivity increases with increase in B;

(ii) the number of turns N of the coil — the sensitivity increases with increase in N.

Let the wire of length L be wound into a circular coil of N turns, each of radius r. Then,

$$\begin{aligned}\text N \times 2\pi\text r = \text L \\[1em] \text r = \frac{\text L}{2\pi\text N}\end{aligned}$$

Area of each turn,

$$\begin{aligned}\text A = \pi\text r^2 \\[1em] = \pi\left(\frac{\text L}{2\pi\text N}\right)^2 \\[1em] \text A = \frac{\text L^2}{4\pi\text N^2}\end{aligned}$$

The torque on the coil in the magnetic field B is τ = NIAB sin θ, which is maximum for θ = 90°. Hence,

$$\begin{aligned}\tau = \text{NIAB} \\[1em] = \text{NIB} \times \frac{\text L^2}{4\pi\text N^2} \\[1em] \tau = \frac{\text L^2\text{IB}}{4\pi\text N}\end{aligned}$$

As $\tau \propto \frac{1}{\text N}$, the torque is greatest for the least possible number of turns, N = 1, i.e., for a coil of 'single' turn. Then,

$\tau_\text{max} = \left(\frac{1}{4\pi}\right)\text L^2\text{IB}$

Hence, the torque is maximum when a coil of single turn is formed and the maximum torque is $\tau = \left(\frac{1}{4\pi}\right)\text L^2\text{IB}$.

A galvanometer is not used as such as an ammeter because :

(i) the resistance of its coil is not very small, so on connecting it in series it would appreciably change the current in the circuit;

(ii) it can measure only a limited current corresponding to the maximum possible deflection of its coil; a larger current may damage it.

A low resistance, called the 'shunt', is connected in parallel with the coil of the galvanometer. The value of the shunt depends on the required range I of the ammeter,

$\text S = \left(\frac{\text I_\text g}{\text I - \text I_\text g}\right)\text G$

where G is the resistance of the galvanometer and I_g the current for full-scale deflection.

An ammeter is connected in series in the circuit. Its resistance must be very small compared with the other resistances of the circuit, so that it does not appreciably change the circuit resistance and hence the current which it measures. An ideal ammeter has zero resistance.

A voltmeter is connected in parallel across the resistance whose potential difference is to be measured. Its resistance must be very high so that it draws negligible current from the circuit and the potential difference to be measured is not changed. An ideal voltmeter has infinite resistance.

No. Neither instrument reads exactly the original value.

Explanation:
When an ammeter is put in a circuit, its own small resistance is introduced into the circuit. Hence, the current recorded is slightly less than the actual current.

When a voltmeter is placed across a part of the circuit, it draws some current. Hence, the voltage drop recorded is slightly less than the original value.

An ammeter is connected in series so that the whole of the current in the circuit passes through it.

A voltmeter measures the potential difference between two points of a circuit, so it is connected in parallel to those points. Due to its high resistance it draws minimum current, so that the potential difference to be measured is not affected materially.

An ammeter has a very low resistance. If it is connected in parallel, it will draw almost the entire current of the circuit and may be damaged.

A voltmeter has a very high resistance. If it is connected in series, it will reduce the current in the circuit to almost zero.

We can increase the range of a given ammeter, but we cannot decrease it.

Explanation:
The range is increased by connecting a proper shunt S in parallel with the ammeter, given by $\text S = \left(\frac{\text I_\text g}{\text I - \text I_\text g}\right)\text G$, where G is the resistance of the ammeter, I_g its present range and I the new (increased) range. A shunt can only divert a part of the current, so the range cannot be decreased.

Yes.

Explanation:
The range of a voltmeter is increased by connecting an additional resistance in series with it. The range is decreased by reducing the total resistance (G + R) of the voltmeter, i.e., by connecting a suitable resistance in parallel with its series resistance R.

Note: The textbook says the range is decreased by connecting a resistance in parallel with the voltmeter. A resistance across the voltmeter terminals leaves the full-scale reading I_g(G + R) unchanged; it must be connected across the series resistance R.

The shunt of the milliammeter will be of larger resistance.

Explanation:
The shunt is given by $\text S = \left(\frac{\text I_\text g}{\text I - \text I_\text g}\right)\text G$. For identical galvanometers, G and I_g are the same, so a smaller range I needs a larger S. In the milliammeter, which has the smaller range, a larger part of the main current passes through the galvanometer coil, so its shunt has the larger resistance.

The ammeter carries the same current I as the wire, but its resistance R_A is very small compared with the resistance R of the wire (R_A << R). Hence, the rate of dissipation of energy I²R_A in the ammeter is very small compared with the rate I²R in the wire.

The voltmeter has the same potential difference V across it as the wire, but its resistance R_V is very high (R_V >> R). Hence, the rate of dissipation of energy V²/R_V in the voltmeter is very small compared with the rate V²/R in the wire.

Therefore, the wire becomes hot but the ammeter and the voltmeter do not.

Let the rectangular coil PQRS have N turns, with vertical sides PQ = RS = l and horizontal sides QR = SP = b, so that the area of each turn is A = l × b. It is suspended in a radial magnetic field of flux density B, produced by cylindrically-cut pole-pieces and a soft-iron core, and a current I passes through it.

Forces on the sides : The vertical sides PQ and RS are always perpendicular to the field. The force on each turn of either side is IBl, so for N turns,

$\text F_1 = \text F_2 = \text{NIBl}$

These forces are equal and opposite, act perpendicular to the plane of the coil and have different lines of action, so they form a deflecting couple. The forces on the sides QR and SP are equal and opposite along the same line, so they cancel.

Torque : If the normal to the plane of the coil makes an angle θ with the field, the perpendicular distance between F₁ and F₂ is b sin θ. Hence,

$$\begin{aligned}\tau = \text{NIBl} \times \text b\sin\theta \\[1em] \tau = \text{NIAB}\sin\theta\end{aligned}$$

In a radial field the plane of the coil remains parallel to the lines of force in every position, i.e., θ = 90° and sin θ = 1. Therefore,

$\tau = \text{NIAB}$

Hence, the torque on the coil in a radial magnetic field is τ = NIAB, independent of the orientation of the coil.

Principle : When an electric current flows in a coil placed in a magnetic field, a deflecting torque acts upon the coil whose magnitude depends upon the strength of the current. From the deflection of the coil, the strength of the current can be computed.

Construction :

A suspended-coil galvanometer consists of a rectangular coil of a large number of turns of fine insulated copper wire wound on a light, non-magnetic aluminium frame. The coil is suspended by a thin phosphor-bronze strip from a torsion head connected to the terminal screw T₁; its lower end is connected to a fine, loosely wound spring joined to the terminal screw T₂. A small light mirror attached to the strip measures the deflection by a lamp and scale arrangement. The coil hangs between the concave cylindrical pole-pieces N and S of a powerful horse-shoe magnet, and a soft-iron cylindrical core is fixed within the coil without touching it.

Current is proportional to deflection : Let the coil have N turns, each of area A, and carry a current I in a field B. The pole-pieces and the core make the field radial, so the plane of the coil is always parallel to the field (θ = 90°) and the deflecting torque is NIAB.

As the coil rotates, the suspension strip is twisted and a restoring torque cφ acts on it, where c is the torsional constant of the strip and φ the deflection. In equilibrium,

$$\begin{aligned}\text{NIAB} = \text c\phi \\[1em] \text I = \left(\frac{\text c}{\text{NAB}}\right)\phi \\[1em] \text I = \text K\phi\end{aligned}$$

where $\text K = \frac{\text c}{\text{NAB}}$ is a constant, called the reduction factor of the galvanometer.

Hence, I ∝ φ, i.e., in a radial field the current in the coil is directly proportional to the deflection of the coil.

Working : The base of the galvanometer is first made horizontal with the levelling screws so that the coil can rotate freely. Light from a lamp falls on the mirror attached to the strip and a spot of reflected light is obtained on a scale. When the current I passes through the coil, a deflecting torque NIAB acts on it in the radial field, and the coil (with the mirror) rotates. The twist in the strip produces a restoring torque cφ, and the coil comes to rest when

$$\begin{aligned}\text{NIAB} = \text c\phi \\[1em] \text I = \left(\frac{\text c}{\text{NAB}}\right)\phi\end{aligned}$$

If the scale is at a distance D from the mirror and the spot of light is displaced through d, then for small φ, $2\phi = \frac{\text d}{\text D}$, i.e., $\phi = \frac{\text d}{2\text D}$. As D is kept constant, I ∝ φ ∝ d, so the current is measured from the displacement of the spot.

Increasing the sensitivity : The current sensitivity of the galvanometer is

$\frac{\phi}{\text I} = \frac{\text{NAB}}{\text c}$

It can be increased by :

(i) increasing B, by using a very strong horse-shoe magnet and placing a soft-iron core within the coil;

(ii) decreasing c, by using a long and fine suspension strip of phosphor-bronze (or quartz);

(iii) increasing N and A, but only up to a limit, otherwise the resistance and the mass of the coil increase undesirably.

A galvanometer has an appreciable resistance and gives full-scale deflection for a very small current. If a heavy current flows through it, its pointer may strike the stop pin and break, or its coil may burn. A small resistance S, the shunt, is therefore connected in parallel with it. Most of the current passes through the shunt and only a small part through the coil, and the combined resistance becomes small, so the circuit current is not appreciably changed.

Derivation : Let G be the resistance of the galvanometer, I the main current, I_g the current through the galvanometer and (I − I_g) the current through the shunt S. As G and S are in parallel, the potential difference across them is the same,

$$\begin{aligned}\text I_\text g \times \text G = (\text I - \text I_\text g) \times \text S \\[1em] \text I_\text g\text G + \text I_\text g\text S = \text{IS} \\[1em] \text I_\text g(\text S + \text G) = \text{IS} \\[1em] \text I_\text g = \frac{\text S}{\text S + \text G}\text I\end{aligned}$$

Hence, $\text I_\text g = \frac{\text S}{\text S + \text G}\text I$, i.e., only the $\frac{\text S}{\text S + \text G}$th part of the main current flows through the galvanometer.

Let G be the resistance of the galvanometer and I_g the current that produces full-scale deflection.

(i) Conversion into an ammeter of range 0 to I :

A small resistance S (shunt) is connected in parallel with the galvanometer.

A current I_g passes through the galvanometer and (I − I_g) through the shunt. As they are in parallel,

$$\begin{aligned}\text I_\text g\text G = (\text I - \text I_\text g)\text S \\[1em] \text S = \left(\frac{\text I_\text g}{\text I - \text I_\text g}\right)\text G\end{aligned}$$

With this shunt, full-scale deflection corresponds to the current I, and the resistance of the ammeter, $\frac{\text{GS}}{\text G + \text S}$, is very small.

(ii) Conversion into a voltmeter of range 0 to V :

A high resistance R is connected in series with the galvanometer.

When the potential difference V is applied, the current through the galvanometer is I_g,

$$\begin{aligned}\text I_\text g = \frac{\text V}{\text G + \text R} \\[1em] \text R = \frac{\text V}{\text I_\text g} - \text G\end{aligned}$$

With this series resistance, full-scale deflection corresponds to V volt, and the resistance of the voltmeter, (G + R), is very high.

(i) Incorrect.

The range of an ammeter is increased by connecting a small resistance (shunt) in parallel with it, so that the extra current bypasses the meter. A high resistance in series does not increase the range; it only reduces the current in the circuit.

(ii) Correct.

Let I_g be the current for full-scale deflection. For the range 0–1 V,

$\text I_\text g(\text G + \text R) = 1 \qquad ...(1)$

If R' is the additional series resistance for the range 0–2 V,

$\text I_\text g(\text G + \text R + \text R') = 2 \qquad ...(2)$

Dividing Eq. (2) by Eq. (1),

$$\begin{aligned}\frac{\text G + \text R + \text R'}{\text G + \text R} = 2 \\[1em] \text G + \text R + \text R' = 2(\text G + \text R) \\[1em] \text R' = \text G + \text R\end{aligned}$$

So an additional resistance (R + G) in series with R doubles the range, as the student did.

(iii) Correct.

Here G = 72 Ω and only 10% of the total current passes through the galvanometer, so $\frac{\text I_\text g}{\text I} = \frac{1}{10}$. For a shunt S,

$$\begin{aligned}\frac{\text I_\text g}{\text I} = \frac{\text S}{\text G + \text S} \\[1em] \frac{1}{10} = \frac{\text S}{72 + \text S} \\[1em] 72 + \text S = 10\text S \\[1em] 9\text S = 72 \\[1em] \text S = 8\ \Omega\end{aligned}$$

So the 8 Ω resistance connected in parallel by the student is the required shunt.

Let a rectangular coil PQRS of N turns, with sides PQ = RS = l and QR = SP = b, carry a current I in a uniform magnetic field $\vec{\text B}$. The area of the coil is A = l × b.

Forces on the sides : The force on a current-carrying side is $\vec{\text F} = \text I\vec{\text l} \times \vec{\text B}$. The sides PQ and RS are always perpendicular to $\vec{\text B}$, so the forces on them are

$\text F_1 = \text F_2 = \text{NIBl}$

These forces are equal, parallel and opposite but have different lines of action, so they form a deflecting couple. The forces F₃ and F₄ on the sides QR and SP are equal and opposite and act along the same line, so they cancel. Hence, the net force on the coil is zero and only a torque acts on it.

(a) In terms of the angle θ which the normal to the plane of the coil makes with the field :

The perpendicular distance between the lines of action of F₁ and F₂ is b sin θ. Therefore,

$$\begin{aligned}\tau = \text F_1 \times \text b\sin\theta \\[1em] = \text{NIBl} \times \text b\sin\theta \\[1em] \tau = \text{NIAB}\sin\theta \qquad ...(i)\end{aligned}$$

With m = NIA as the magnetic moment of the coil, τ = mB sin θ, or in vector form $\vec{\tau} = \vec{\text m} \times \vec{\text B}$.

(b) In terms of the angle α which the plane of the coil makes with the field :

The normal is perpendicular to the plane of the coil, so θ = 90° − α. Putting this in Eq. (i),

$$\begin{aligned}\tau = \text{NIAB}\sin\left(90^\circ - \alpha\right) \\[1em] \tau = \text{NIAB}\cos\alpha \qquad ...(ii)\end{aligned}$$

Special cases :

• When the plane of the coil is parallel to the field (θ = 90°, α = 0°), the torque is maximum, τ_max = NIAB.
• When the plane of the coil is perpendicular to the field (θ = 0°, α = 90°), the torque is zero.

Hence, τ = NIAB sin θ in terms of the angle with the normal, and τ = NIAB cos α in terms of the angle with the plane of the coil.

Radial field : A radial magnetic field is one in which the lines of force are directed along the radii of the cylindrical soft-iron core, so that the plane of the coil remains parallel to the field in every position of the coil (θ = 90°). It is obtained by using cylindrically-cut (concave) pole-pieces and a soft-iron core within the coil.

Sensitivity : A galvanometer is said to be sensitive if a small current passed through it produces a sufficiently large deflection.

• The current sensitivity is the deflection produced in the galvanometer when a unit current flows through it, $\frac{\phi}{\text I}$. Its unit is radian/ampere.
• The voltage sensitivity is the deflection produced in the galvanometer when a unit voltage is applied across its coil, $\frac{\phi}{\text V}$. Its unit is radian/volt.

Factors affecting the current sensitivity : In a radial field a current I through a coil of N turns, each of area A, produces a deflecting torque NIAB. It is balanced by the restoring torque cφ of the suspension strip, where c is its torsional constant. In equilibrium,

$$\begin{aligned}\text{NIAB} = \text c\phi \\[1em] \frac{\phi}{\text I} = \frac{\text{NAB}}{\text c}\end{aligned}$$

Hence, the current sensitivity depends on :

(i) the number of turns N — increasing N increases the sensitivity, but N cannot be increased beyond a limit, otherwise the resistance and the mass of the coil increase undesirably;

(ii) the area A of the coil — a larger area increases the sensitivity, but it is limited for the same reason;

(iii) the magnetic field B — B is made as large as possible by using a very strong horse-shoe magnet and placing a soft-iron core within the coil, which concentrates the lines of force;

(iv) the torsional constant c — c is made as small as possible by using a long and fine suspension strip of phosphor-bronze (or a silvered quartz fibre).

Construction :

The suspended-coil galvanometer consists of the following parts :

(i) Coil — a rectangular (or circular) coil of a large number of turns of fine insulated copper wire wound on a light, non-magnetic metallic (aluminium) frame.

(ii) Suspension — the coil is suspended by a thin phosphor-bronze strip from a torsion head connected to a terminal screw T₁. Its lower end is connected to a fine, loosely wound spring joined to another terminal screw T₂.

(iii) Mirror — a small light mirror attached to the lower portion of the strip rotates with it, and its deflection is measured by a lamp and scale arrangement.

(iv) Magnet and core — the coil hangs between the pole-pieces of a powerful permanent horse-shoe magnet NS, which are cut concave cylindrical. A soft-iron cylindrical core is fixed within the coil without touching it.

(v) Case — the whole arrangement is enclosed in a non-magnetic box with a glass front, and three levelling screws are provided at its base.

Why the poles are concave : The concave (cylindrically-cut) pole-pieces, together with the soft-iron core, make the magnetic field radial. The plane of the coil then remains parallel to the field in every position, so the deflecting torque is always NIAB. In equilibrium NIAB = cφ, giving I ∝ φ, so the deflection is directly proportional to the current and the scale is uniform. The core also concentrates the lines of force and makes the field strong.

Why phosphor-bronze strip is used :

(i) Its torsional constant (torsion couple per unit twist) is small, which makes the galvanometer sensitive.

(ii) Its tensile strength is large enough to support the coil even when the strip is very fine.

(iii) It is a good conductor of electricity, so it also serves to lead the current into the coil.

(iv) It does not oxidise (is rust proof) and has very less elastic after effect.

(i) Voltmeter C (20,000 Ω) is the best.

Explanation:
A voltmeter is connected in parallel and should draw as little current as possible from the circuit. C has the highest resistance, so it draws the least current and gives the most accurate reading.

(ii) Given,

• Full-scale current, I_g = 25 mA = 25 × 10^-3 A
• Resistance of the milliammeter, G = 10 Ω
• Required range, V = 25 V

The resistance to be connected in series is

$$\begin{aligned}\text R = \frac{\text V}{\text I_\text g} - \text G \\[1em] = \frac{25}{25 \times 10^{-3}} - 10 \\[1em] = 1000 - 10 \\[1em] \text R = 990\ \Omega\end{aligned}$$

So, a resistance of 990 Ω should be connected in series with it.

(iii) Here the resistance of the voltmeter is G = R Ω and n = 3. The resistance to be connected in series is

$$\begin{aligned}\text R_\text s = (\text n - 1)\text G \\[1em] = (3 - 1)\text R \\[1em] \text R_\text s = 2\text R\end{aligned}$$

So, a resistance of 2R Ω should be connected in series with it.

(iv) Infinite.

Explanation:
An ideal voltmeter draws no current from the circuit.

Given,

• Magnetic moment of the coil, m = 2.5 × 10^-8 J T^-1
• Torque on the coil, τ = 7.5 × 10^-9 J

The face (plane) of the coil is parallel to the field, so the normal to the coil makes an angle θ = 90° with the field. The coil therefore experiences the maximum torque.

The torque on a coil of magnetic moment m in a uniform magnetic field B is

$\tau = \text{mB} \sin\theta$

With θ = 90°,

$$\begin{aligned}\tau = \text{mB} \sin 90^\circ \\[1em] \tau = \text{mB} \\[1em] \text B = \frac{\tau}{\text m} \\[1em] \text B = \frac{7.5 \times 10^{-9}}{2.5 \times 10^{-8}} \\[1em] \text B = 0.30\ \text T\end{aligned}$$

Hence, the magnitude of the magnetic field is 0.30 T.

Given,

• Number of turns, N = 20
• Area of the coil, A = 10 cm × 10 cm = 100 cm² = 100 × 10^-4 m²
• Current, I = 12 A
• Magnetic field, B = 0.80 T
• Angle between the normal to the coil and the field, θ = 30°

The torque on a coil of N turns carrying current I in a uniform magnetic field B is

$\tau = \text{NIAB} \sin\theta$

Substituting the values,

$$\begin{aligned}\tau = 20 \times 12 \times \left(100 \times 10^{-4}\right) \times 0.80 \times \sin 30^\circ \\[1em] = 240 \times 10^{-2} \times 0.80 \times \frac{1}{2} \\[1em] \tau = 0.96\ \text{N m}\end{aligned}$$

Hence, the magnitude of the torque experienced by the coil is 0.96 N m.

Given,

• Area of the loop, A = 5 m²
• Number of turns, N = 50
• Current, I = 1 A
• Magnetic field, B = 0.1 T
• Angle between the field and the plane of the coil = 30°

In τ = NIAB sin θ, the angle θ is measured between the normal to the plane of the coil and the field. As the field makes 30° with the plane of the coil,

$$\begin{aligned}\theta = 90^\circ - 30^\circ \\[1em] \theta = 60^\circ\end{aligned}$$

Substituting the values in τ = NIAB sin θ,

$$\begin{aligned}\tau = 50 \times 1 \times 5 \times 0.1 \times \sin 60^\circ \\[1em] = 25 \times 0.866 \\[1em] \tau = 21.65\ \text{N m}\end{aligned}$$

Hence, the torque experienced by the coil is 21.65 N m.

Note: The question does not say from what the 30° is measured. The printed answer, 21.65 N m, takes it as the angle with the plane of the coil. Measured from the normal, the torque would be 12.5 N m.

Given,

• Radius of the coil, r = 6.0 cm = 6.0 × 10^-2 m
• Number of turns, N = 25
• Current, I = 10 A
• Magnetic field, B = 1.2 T

The field lines lie in the plane of the coil, so the normal to the coil is perpendicular to the field, θ = 90°.

Area of the coil,

$$\begin{aligned}\text A = \pi\text r^2 \\[1em] = 3.14 \times \left(6.0 \times 10^{-2}\right)^2 \\[1em] = 3.14 \times 36 \times 10^{-4} \\[1em] \text A = 1.13 \times 10^{-2}\ \text m^2\end{aligned}$$

The torque on the coil is τ = NIAB sin θ. With θ = 90°,

$$\begin{aligned}\tau = 25 \times 10 \times \left(1.13 \times 10^{-2}\right) \times 1.2 \times \sin 90^\circ \\[1em] = 250 \times 1.13 \times 10^{-2} \times 1.2 \\[1em] = 3.39 \\[1em] \tau \approx 3.4\ \text{N m}\end{aligned}$$

Hence, the torque acting on the coil is 3.4 N m.

Given,

• Area of the coil, A = 8.0 cm × 6.0 cm = 48 cm² = 48 × 10^-4 m²
• Number of turns, N = 2000
• Current, I = 0.20 A
• Magnetic field, B = 0.20 T, along the positive X-axis

(a) Net force on the coil :

In a uniform magnetic field, the forces on each pair of opposite sides of the coil are equal in magnitude and opposite in direction, and they cancel each other. Hence, the net force on the coil is zero.

(b) Maximum torque :

The torque on the coil is τ = NIAB sin θ, where θ is the angle between the normal to the plane of the coil and the field. It is maximum when θ = 90°, so

$$\begin{aligned}\tau_\text{max} = \text{NIAB} \\[1em] = 2000 \times 0.20 \times \left(48 \times 10^{-4}\right) \times 0.20 \\[1em] = 400 \times 48 \times 10^{-4} \times 0.20 \\[1em] \tau_\text{max} = 0.384\ \text{N m}\end{aligned}$$

θ = 90° means that the plane of the coil is parallel to the field.

The maximum torque is 0.384 N m, when the X-axis lies in the plane of the coil.

(c) Orientation for zero torque :

τ = 0 when sin θ = 0, that is, θ = 0° or 180°. The normal to the coil is then along the X-axis, so the plane of the coil is perpendicular to the field. The torque is zero when the coil lies in the Y-Z plane.

Stable equilibrium : when θ = 0°, that is, the area vector $\vec{\text A} \space$(and the magnetic moment $\vec{\text m}$) is parallel to $\vec{\text B}$. If the coil is turned slightly from this position, the torque brings it back.

Unstable equilibrium : when θ = 180°, that is, $\vec{\text A} \space$is anti-parallel to $\vec{\text B}$. If the coil is turned slightly from this position, the torque turns it further away, until $\vec{\text A} \space$becomes parallel to $\vec{\text B}$.

Given,

• Area of the coil, A = 5 cm² = 5 × 10^-4 m²
• Number of turns, N = 100
• Magnetic field, B = 1.5 T
• Current, I = 0.2 A

Magnetic dipole moment :

The magnitude of the magnetic moment of a coil is m = NIA. Hence,

$$\begin{aligned}\text m = 100 \times 0.2 \times \left(5 \times 10^{-4}\right) \\[1em] \text m = 1.0 \times 10^{-2}\ \text{A m}^2\end{aligned}$$

Maximum torque :

From τ = mB sin θ, the torque is maximum when θ = 90°. Hence,

$$\begin{aligned}\tau_\text{max} = \text{mB} \\[1em] = \left(1.0 \times 10^{-2}\right) \times 1.5 \\[1em] \tau_\text{max} = 1.5 \times 10^{-2}\ \text{N m}\end{aligned}$$

Hence, the magnetic dipole moment of the coil is 1.0 × 10^-2 A m² and the maximum torque is 1.5 × 10^-2 N m.

Given,

• Magnetic field, B = 3000 G = 3000 × 10^-4 T = 0.30 T, along the positive Z-axis, so $\vec{\text B} = 0.30\ \hat{\text k}$ T
• Area of the loop, A = 10 cm × 5 cm = 50 cm² = 50 × 10^-4 m²
• Current, I = 12 A

The torque on a current-loop in a uniform magnetic field is

$\vec\tau = \text I\vec{\text A} \times \vec{\text B}$

where the area vector $\vec{\text A} \space$is normal to the plane of the loop, its sense given by the right-hand rule from the direction of the current.

Case (a) :

The loop lies in the Y-Z plane. From the sense of the current, $\vec{\text A} \space$is along the positive X-axis, so $\vec{\text A} = 50 \times 10^{-4}\ \hat{\text i}$ m².

$$\begin{aligned}\vec\tau = 12 \times \left(50 \times 10^{-4}\ \hat{\text i}\right) \times \left(0.30\ \hat{\text k}\right) \\[1em] = 1.8 \times 10^{-2}\left(\hat{\text i} \times \hat{\text k}\right) \\[1em] \vec\tau = -1.8 \times 10^{-2}\ \hat{\text j}\ \text{N m}\end{aligned}$$

The torque is 1.8 × 10^-2 N m along the negative Y-axis.

Case (b) :

The loop again lies in the Y-Z plane with the current in the same sense, so $\vec{\text A} \space$is again along the positive X-axis and the angle between $\vec{\text A} \space$and $\vec{\text B} \space$is 90°. The torque is the same as in case (a):

$\vec\tau = -1.8 \times 10^{-2}\ \hat{\text j}\ \text{N m}$

The torque is 1.8 × 10^-2 N m along the negative Y-axis.

Case (c) :

The loop lies in the Z-X plane. From the sense of the current, $\vec{\text A} \space$is along the negative Y-axis, so $\vec{\text A} = -50 \times 10^{-4}\ \hat{\text j}$ m².

$$\begin{aligned}\vec\tau = 12 \times \left(-50 \times 10^{-4}\ \hat{\text j}\right) \times \left(0.30\ \hat{\text k}\right) \\[1em] = -1.8 \times 10^{-2}\left(\hat{\text j} \times \hat{\text k}\right) \\[1em] \vec\tau = -1.8 \times 10^{-2}\ \hat{\text i}\ \text{N m}\end{aligned}$$

The torque is 1.8 × 10^-2 N m along the negative X-axis.

Case (d) :

The loop lies in the X-Y plane, so $\vec{\text A} \space$is along the Z-axis, parallel to $\vec{\text B}$. The angle between $\vec{\text A} \space$and $\vec{\text B} \space$is 0°, so

$$\begin{aligned}\tau = \text{IAB} \sin 0^\circ \\[1em] \tau = 0\end{aligned}$$

Hence, the torque is 1.8 × 10^-2 N m along the negative Y-axis in cases (a) and (b), 1.8 × 10^-2 N m along the negative X-axis in case (c), and zero in case (d).

Given,

• Initial number of turns, N = 24
• Initial resistance of the coil, R = 12 Ω
• New resistance of the coil, R' = 20 Ω

Increase in the number of turns :

The current-sensitivity of a moving-coil galvanometer is

$\text S_\text c = \frac{\phi}{\text I} = \frac{\text{NAB}}{\text c}$

where c is the torsional constant of the suspension. With A, B and c unchanged, S_c ∝ N. To increase S_c by 25%, N must also be increased by 25%:

$$\begin{aligned}\text N' = 1.25 \times \text N \\[1em] \text N' = 1.25 \times 24 \\[1em] \text N' = 30\end{aligned}$$

So the number of turns must be increased by 30 − 24 = 6.

Effect on the voltage-sensitivity :

The voltage-sensitivity is

$$\begin{aligned}\text S_\text V = \frac{\phi}{\text V} \\[1em] = \frac{\phi}{\text{IR}} \\[1em] = \frac{\text{NAB}}{\text{cR}} \\[1em] = \frac{\text N}{\text R}\left(\frac{\text{AB}}{\text c}\right)\end{aligned}$$

Original voltage-sensitivity, with N = 24 and R = 12 Ω :

$$\begin{aligned}\text S_\text V = \frac{24}{12}\left(\frac{\text{AB}}{\text c}\right) \\[1em] \text S_\text V = 2.0\ \frac{\text{AB}}{\text c}\end{aligned}$$

New voltage-sensitivity, with N' = 30 and R' = 20 Ω :

$$\begin{aligned}\text S_\text V' = \frac{30}{20}\left(\frac{\text{AB}}{\text c}\right) \\[1em] \text S_\text V' = 1.5\ \frac{\text{AB}}{\text c}\end{aligned}$$

Therefore,

$$\begin{aligned}\frac{\text S_\text V'}{\text S_\text V} = \frac{1.5}{2.0} \\[1em] \frac{\text S_\text V'}{\text S_\text V} = 0.75\end{aligned}$$

Hence, the number of turns should be increased by 6, from 24 to 30. The voltage-sensitivity then falls to 0.75 of its original value, a decrease of 25%.

Given,

• New current-sensitivity, S_c' = 1.5 S_c (an increase of 50%)
• New resistance, R' = 2R

The voltage-sensitivity of a galvanometer is related to its current-sensitivity by

$\text S_\text V = \frac{\text S_\text c}{\text R}$

So the new voltage-sensitivity is

$$\begin{aligned}\text S_\text V' = \frac{\text S_\text c'}{\text R'} \\[1em] = \frac{1.5\ \text S_\text c}{2\text R} \\[1em] = 0.75\ \frac{\text S_\text c}{\text R} \\[1em] \text S_\text V' = 0.75\ \text S_\text V\end{aligned}$$

Hence, the voltage-sensitivity changes by a factor of 0.75, that is, it is reduced to 0.75 of its previous value.

Given,

• Resistance of the galvanometer, G = 100 Ω
• PD for full-scale deflection, V_g = 100 mV = 0.1 V
• Range of the ammeter, I = 5 A

The current for full-scale deflection in the galvanometer is

$$\begin{aligned}\text I_\text g = \frac{\text V_\text g}{\text G} \\[1em] \text I_\text g = \frac{0.1}{100} \\[1em] \text I_\text g = 1 \times 10^{-3}\ \text A\end{aligned}$$

The shunt S required to convert the galvanometer into an ammeter of range I is

$\text S = \left(\frac{\text I_\text g}{\text I - \text I_\text g}\right)\text G$

Substituting the values,

$$\begin{aligned}\text S = \frac{1 \times 10^{-3}}{5 - 0.001} \times 100 \\[1em] = \frac{0.1}{4.999} \\[1em] \text S \approx 0.02\ \Omega\end{aligned}$$

Hence, a shunt of 0.02 Ω is required.

Given,

• Resistance of the galvanometer, G = 99 Ω
• Current for full-scale deflection, I_g = 50 mA = 0.05 A
• Range of the ammeter, I = 5 A

Shunt required :

$\text S = \left(\frac{\text I_\text g}{\text I - \text I_\text g}\right)\text G$

Substituting the values,

$$\begin{aligned}\text S = \frac{0.05}{5 - 0.05} \times 99 \\[1em] = \frac{0.05 \times 99}{4.95} \\[1em] = \frac{4.95}{4.95} \\[1em] \text S = 1.0\ \Omega\end{aligned}$$

Resistance of the ammeter :

The galvanometer and the shunt are in parallel, so the resistance of the ammeter is

$$\begin{aligned}\text R_\text A = \frac{\text{GS}}{\text G + \text S} \\[1em] = \frac{99 \times 1.0}{99 + 1.0} \\[1em] \text R_\text A = 0.99\ \Omega\end{aligned}$$

Hence, the shunt required is 1.0 Ω and the resistance of the ammeter is 0.99 Ω.

Given,

• Resistance of the galvanometer, G = 30 Ω
• Number of divisions on the scale = 100
• Current per division = 0.2 mA
• Current to be measured, I = 5 A

The current for full-scale deflection (100 divisions) is

$$\begin{aligned}\text I_\text g = 0.2\ \text{mA/div} \times 100\ \text{div} \\[1em] = 20\ \text{mA} \\[1em] \text I_\text g = 0.02\ \text A\end{aligned}$$

The required shunt is given by

$\text S = \left(\frac{\text I_\text g}{\text I - \text I_\text g}\right)\text G$

Substituting the values,

$$\begin{aligned}\text S = \frac{0.02}{5 - 0.02} \times 30 \\[1em] = \frac{0.6}{4.98} \\[1em] \text S \approx 0.12\ \Omega\end{aligned}$$

Hence, a shunt of 0.12 Ω will be required.

Given,

• Resistance of the galvanometer, G = 80 Ω
• Deflection = 1 division per mA
• Number of divisions on the scale = 50
• Shunt, S = 2.5 Ω

The current required for full-scale (50 divisions) deflection is

$$\begin{aligned}\text I_\text g = \frac{50\ \text{div}}{1\ \text{div/mA}} \\[1em] \text I_\text g = 50\ \text{mA}\end{aligned}$$

Let I be the maximum current the shunted galvanometer can measure. Of this, I_g passes through the galvanometer and (I − I_g) through the shunt. As G and S are in parallel, the PD across them is the same:

$$\begin{aligned}\text I_\text g \times \text G = \left(\text I - \text I_\text g\right) \times \text S \\[1em] \text I = \left(\frac{\text S + \text G}{\text S}\right)\text I_\text g\end{aligned}$$

Substituting the values,

$$\begin{aligned}\text I = \left(\frac{2.5 + 80}{2.5}\right) \times 50\ \text{mA} \\[1em] = 33 \times 50\ \text{mA} \\[1em] = 1650\ \text{mA} \\[1em] \text I = 1.65\ \text A\end{aligned}$$

Hence, the galvanometer can measure a maximum current of 1.65 A.

Given,

• Resistance of the galvanometer, G = 100 Ω
• Current for full-scale deflection, I_g = 5 × 10^-4 A
• Range of the voltmeter, V = 5 V

To convert a galvanometer into a voltmeter, a high resistance R is connected in series with it, such that

$\text R = \frac{\text V}{\text I_\text g} - \text G$

Substituting the values,

$$\begin{aligned}\text R = \frac{5}{5 \times 10^{-4}} - 100 \\[1em] = 10000 - 100 \\[1em] \text R = 9900\ \Omega\end{aligned}$$

Hence, the galvanometer is converted into a voltmeter of range 0 to 5 V by connecting a resistance of 9900 Ω in series with it.

Given,

• Resistance of the galvanometer, G = 50 Ω
• Number of divisions on the scale = 25
• Current per division = 4.0 × 10^-4 A
• Range of the voltmeter, V = 2.5 V

The current for full-scale (25 divisions) deflection is

$$\begin{aligned}\text I_\text g = 4.0 \times 10^{-4} \times 25 \\[1em] \text I_\text g = 1.0 \times 10^{-2}\ \text A\end{aligned}$$

The resistance R to be connected in series is given by

$\text R = \frac{\text V}{\text I_\text g} - \text G$

Substituting the values,

$$\begin{aligned}\text R = \frac{2.5}{1.0 \times 10^{-2}} - 50 \\[1em] = 250 - 50 \\[1em] \text R = 200\ \Omega\end{aligned}$$

Hence, a resistance of 200 Ω should be connected in series with the galvanometer.

Given,

• Resistance of the voltmeter, G = 180 Ω
• PD for full-scale deflection = 5 V
• New range, V' = 25 V

The current giving full-scale deflection in the voltmeter is

$$\begin{aligned}\text I_\text g = \frac{5}{180} \\[1em] \text I_\text g = \frac{1}{36}\ \text A\end{aligned}$$

Let R be the resistance connected in series with the voltmeter so that it reads up to V' = 25 V. Then

$\text R = \frac{\text V'}{\text I_\text g} - \text G$

Substituting the values,

$$\begin{aligned}\text R = 25 \times 36 - 180 \\[1em] = 900 - 180 \\[1em] \text R = 720\ \Omega\end{aligned}$$

Hence, a resistance of 720 Ω should be connected in series.

Given,

• Resistance of the galvanometer coil, G = 15 Ω
• Current for full-scale deflection, I_g = 2.0 mA = 2.0 × 10^-3 A

(a) Conversion into an ammeter of range 0 to 5.0 A :

A shunt S is connected in parallel with the galvanometer, where

$\text S = \left(\frac{\text I_\text g}{\text I - \text I_\text g}\right)\text G$

With I = 5.0 A,

$$\begin{aligned}\text S = \frac{2.0 \times 10^{-3}}{5.0 - 0.002} \times 15 \\[1em] = \frac{0.030}{4.998} \\[1em] \text S \approx 6.0 \times 10^{-3}\ \Omega\end{aligned}$$

The net resistance of the ammeter (G and S in parallel) is

$$\begin{aligned}\text R_\text A = \frac{\text{GS}}{\text G + \text S} \\[1em] = \frac{15 \times 6.0 \times 10^{-3}}{15 + 6.0 \times 10^{-3}} \\[1em] = \frac{0.090}{15.006} \\[1em] = 5.998 \times 10^{-3} \\[1em] \text R_\text A \approx 6.0 \times 10^{-3}\ \Omega\end{aligned}$$

(b) Conversion into a voltmeter of range 0 to 15.0 V :

A resistance R is connected in series with the galvanometer, where

$\text R = \frac{\text V}{\text I_\text g} - \text G$

With V = 15.0 V,

$$\begin{aligned}\text R = \frac{15.0}{2.0 \times 10^{-3}} - 15 \\[1em] = 7500 - 15 \\[1em] \text R = 7485\ \Omega\end{aligned}$$

The net resistance of the voltmeter (G and R in series) is

$$\begin{aligned}\text R_\text V = \text R + \text G \\[1em] = 7485 + 15 \\[1em] \text R_\text V = 7500\ \Omega\end{aligned}$$

Hence, (a) a shunt of 6.0 × 10^-3 Ω in parallel makes an ammeter of net resistance 6.0 × 10^-3 Ω, and (b) a resistance of 7485 Ω in series makes a voltmeter of net resistance 7500 Ω.

Note: The textbook prints the ammeter's resistance as 0.00599 Ω, which is 5.998 × 10^-3 Ω truncated. Correctly rounded, it is 6.0 × 10^-3 Ω.

Given,

• Reading of the ammeter, I = 2 A
• Reading of the voltmeter, V = 120 V
• Resistance of the voltmeter, R_V = 3000 Ω

The voltmeter is connected in parallel with R, so the PD across R is 120 V. The ammeter is in the main line, so the current of 2 A divides between R and the voltmeter.

When the resistance of the voltmeter is 3000 Ω :

Current through the voltmeter,

$$\begin{aligned}\text I_\text V = \frac{\text V}{\text R_\text V} \\[1em] = \frac{120}{3000} \\[1em] \text I_\text V = 0.04\ \text A\end{aligned}$$

Current through R,

$$\begin{aligned}\text I_\text R = \text I - \text I_\text V \\[1em] = 2 - 0.04 \\[1em] \text I_\text R = 1.96\ \text A\end{aligned}$$

Therefore,

$$\begin{aligned}\text R = \frac{\text V}{\text I_\text R} \\[1em] = \frac{120}{1.96} \\[1em] \text R = 61.2\ \Omega\end{aligned}$$

When the resistance of the voltmeter is infinitely large :

No current flows through the voltmeter, so the whole 2 A flows through R:

$$\begin{aligned}\text R = \frac{\text V}{\text I} \\[1em] = \frac{120}{2} \\[1em] \text R = 60\ \Omega\end{aligned}$$

Hence, R = 61.2 Ω when the resistance of the voltmeter is 3000 Ω, and R = 60 Ω when the resistance of the voltmeter is taken to be infinitely large.

A shunt is a low resistance connected in parallel with the coil of the galvanometer. It carries the major part of the current, so that only the small current I_g that gives full-scale deflection passes through the coil. Thus the shunt protects the coil from a large current and makes the resistance of the ammeter very low, so that the ammeter can measure a large current without appreciably changing the current in the circuit.

Purpose of concave pole-pieces : The pole-pieces of the magnet are cut concave cylindrical and a soft-iron cylindrical core is placed within the coil. This makes the magnetic field radial, so that in every position of the coil its plane remains parallel to the lines of force. The deflecting torque on the coil is then always NIAB, and the current becomes directly proportional to the deflection (I ∝ φ), giving a linear scale.

(i) Conversion into ammeter : A low resistance S, called a shunt, is connected in parallel with the galvanometer G.

$\text S = \left(\frac{\text I_\text g}{\text I - \text I_\text g}\right)\text G$

(ii) Conversion into voltmeter : A high resistance R is connected in series with the galvanometer G.

$\text R = \frac{\text V}{\text I_\text g} - \text G$

Given,

• Resistance of the galvanometer, G = 50 Ω
• Series resistor, R = 10 kΩ = 10⁴ Ω
• Shunt resistor, S = 10 mΩ = 10^-2 Ω

(a) Resistance of the voltmeter :

The resistor is joined in series with the galvanometer, so

$$\begin{aligned}\text R_\text V = \text G + \text R \\[1em] = 50 + 10000 \\[1em] = 10050\ \Omega \\[1em] \text R_\text V \approx 1.0 \times 10^4\ \Omega\end{aligned}$$

(b) Resistance of the ammeter :

The resistor is joined in parallel with the galvanometer, so

$$\begin{aligned}\text R_\text A = \frac{\text{GS}}{\text G + \text S} \\[1em] = \frac{50 \times 0.01}{50 + 0.01} \\[1em] = \frac{0.5}{50.01} \\[1em] = 0.009998\ \Omega \\[1em] \text R_\text A \approx 1.0 \times 10^{-2}\ \Omega\end{aligned}$$

Hence, to two significant figures, the resistance of the voltmeter is 1.0 × 10⁴ Ω (10 kΩ) and that of the ammeter is 1.0 × 10^-2 Ω (10 mΩ).

(i) The number of turns N in the coil.

(ii) The strength of the magnetic field B.

Explanation: The current sensitivity of a moving coil galvanometer is

$$\begin{aligned}\text S_\text c = \frac{\phi}{\text I} \\[1em] \text S_\text c = \frac{\text{NAB}}{\text c}\end{aligned}$$

so it increases with the number of turns N, the area A of the coil and the magnetic field B, and decreases with the torsional constant c of the suspension.

Let a coil of N turns, each of area A, carrying a current I be suspended in a radial magnetic field B. Its magnetic dipole moment, directed along the normal to the plane of the coil, is

$\text M = \text{NIA}$

Deflecting torque : From $\vec{\tau} = \vec{\text M} \times \vec{\text B}$, the magnitude of the torque is

$\tau = \text{MB}\sin\theta \qquad ...(i)$

where θ is the angle between $\vec{\text M} \space$(the normal to the coil) and $\vec{\text B}$.

In a radial field the plane of the coil remains parallel to the lines of force in every position, so the normal to the coil is always perpendicular to the field, i.e., θ = 90° and sin θ = 1. Hence, from Eq. (i),

$$\begin{aligned}\tau = \text{MB} \\[1em] \tau = \text{NIAB}\end{aligned}$$

Restoring torque : As the coil turns through an angle φ, the suspension strip is twisted and exerts a restoring torque cφ, where c is the torsional constant of the strip.

Equilibrium : The coil comes to rest when the two torques balance,

$$\begin{aligned}\text{NIAB} = \text c\phi \\[1em] \text I = \left(\frac{\text c}{\text{NAB}}\right)\phi \\[1em] \text I = \text K\phi\end{aligned}$$

where $\text K = \frac{\text c}{\text{NAB}}$ is a constant, called the reduction factor of the galvanometer.

Hence, I = Kφ, i.e., the current in the coil is directly proportional to its deflection.

Given,

• Number of turns, N = 50
• Area of each turn, A = 0.01 m²
• Current, I = 2 A

The magnetic dipole moment of a current-carrying coil is M = NIA.

$$\begin{aligned}\text M = 50 \times 2 \times 0.01 \\[1em] \text M = 1\ \text A\ \text m^2\end{aligned}$$

Hence, the magnetic dipole moment of the coil is 1 A m².

Given,

• Resistance of the galvanometer, G = 100 Ω
• Number of divisions = 100
• Current sensitivity = 25 μA/division
• Required range, I = 5 A

The current for full-scale deflection is

$$\begin{aligned}\text I_\text g = 25\ \mu\text A/\text{div} \times 100\ \text{div} \\[1em] = 2500\ \mu\text A \\[1em] \text I_\text g = 2.5 \times 10^{-3}\ \text A\end{aligned}$$

To convert the galvanometer into an ammeter, a shunt S is connected in parallel with its coil, where

$\text S = \left(\frac{\text I_\text g}{\text I - \text I_\text g}\right)\text G$

Substituting the values,

$$\begin{aligned}\text S = \frac{2.5 \times 10^{-3}}{5 - 2.5 \times 10^{-3}} \times 100 \\[1em] = \frac{0.25}{4.9975} \\[1em] \text S \approx 0.05\ \Omega\end{aligned}$$

Hence, the galvanometer is converted into an ammeter of range 0 to 5 A by connecting a shunt of 0.05 Ω in parallel with its coil.

a high resistance in series with its coil

Reason — A voltmeter must have a high resistance so that it draws negligible current from the part of the circuit across which it is connected. A high resistance $\text R = \frac{\text V}{\text I_\text g} - \text G$ joined in series with the coil raises its resistance and allows only the full-scale current I_g to flow when the potential difference V is applied.

A radial magnetic field is one in which the magnetic lines of force are directed along the radii of the cylindrical soft-iron core, so that in every position of the coil its plane remains parallel to the lines of force (θ = 90°).

It is produced by cutting the pole-pieces of the horse-shoe magnet concave cylindrical and placing a soft-iron cylindrical core within the coil. In such a field the deflecting torque on the coil is always NIAB.

Ammeter : The first galvanometer G₁ is converted into an ammeter by connecting the low resistance R_L in parallel with it. R_L acts as a shunt and carries most of the current, and the combination has a low resistance.

Voltmeter : The second galvanometer G₂ is converted into a voltmeter by connecting the high resistance R_H in series with it, so that the combination has a high resistance.

Given,

• Resistance of the coil, G = 59 Ω
• Current for full-scale deflection, I_g = 50 mA = 0.05 A
• Required range, I = 3 A

To convert the galvanometer into an ammeter, a shunt S is connected in parallel with its coil, where

$\text S = \left(\frac{\text I_\text g}{\text I - \text I_\text g}\right)\text G$

Substituting the values,

$$\begin{aligned}\text S = \frac{0.05 \times 59}{3 - 0.05} \\[1em] = \frac{2.95}{2.95} \\[1em] \text S = 1\ \Omega\end{aligned}$$

Hence, the galvanometer is converted into an ammeter of range 0 to 3 A by connecting a shunt of 1 Ω in parallel with its coil.

The pole-pieces are made cylindrical (concave), with a soft-iron cylindrical core placed within the coil, to make the magnetic field radial.

In a radial field the plane of the coil remains parallel to the lines of force in every position, so the deflecting torque on the coil is always NIAB. In equilibrium NIAB = cφ, which gives I ∝ φ. Hence the deflection is directly proportional to the current and the galvanometer has a linear (uniform) scale.

Given,

• Area of the loop, A = 5 m²
• Number of turns, N = 50
• Current, I = 1 A
• Magnetic field, B = 0.1 T
• Angle between the field and the plane of the loop = 30°

The torque on a current-carrying coil is

$\tau = \text{NIAB}\sin\theta$

where θ is the angle between the normal to the plane of the coil and the field. Here,

θ = 90° − 30° = 60°

Substituting the values,

$$\begin{aligned}\tau = 50 \times 1 \times 5 \times 0.1 \times \sin 60^\circ \\[1em] = 25 \times 0.866 \\[1em] \tau = 21.65\ \text{N m}\end{aligned}$$

Hence, the torque experienced by the coil is 21.65 N m.

Note: The angle of 30° is taken between the field and the plane of the coil, as the printed answer (21.65 N-m) implies; taken with the normal, the torque would be 12.5 N m.

A moving coil galvanometer is converted into a voltmeter by connecting a suitable high resistance R in series with its coil.

If G is the resistance of the galvanometer, I_g the current for full-scale deflection and V the required range, then

$$\begin{aligned}\text I_\text g = \frac{\text V}{\text G + \text R} \\[1em] \text R = \frac{\text V}{\text I_\text g} - \text G\end{aligned}$$

The resistance of the voltmeter (G + R) is high, so it draws very little current from the circuit.

A moving coil galvanometer is converted into an ammeter by connecting a suitable low resistance S, called a shunt, in parallel with its coil.

If I is the required range and I_g the current for full-scale deflection, then (I − I_g) passes through the shunt. Since G and S are in parallel, the potential difference across them is the same,

$$\begin{aligned}\text I_\text g \times \text G = (\text I - \text I_\text g) \times \text S \\[1em] \text S = \left(\frac{\text I_\text g}{\text I - \text I_\text g}\right)\text G\end{aligned}$$

The resistance of the ammeter so formed is very low.

(i) R₂ is connected in parallel with G

(ii) R₁ is connected in series with G

Reason — An ammeter is made by joining a low resistance (shunt) in parallel with the galvanometer, so the low resistance R₂ = 0.1 Ω is connected in parallel with G. A voltmeter is made by joining a high resistance in series with the galvanometer, so the high resistance R₁ = 5000 Ω is connected in series with G.

Given,

• Resistance of the galvanometer, G = 55 Ω
• Current for full-scale deflection, I_g = 250 mA = 0.25 A
• Required range, I = 3 A

To convert the galvanometer into an ammeter, a shunt S is connected in parallel with its coil, where

$\text S = \left(\frac{\text I_\text g}{\text I - \text I_\text g}\right)\text G$

Substituting the values,

$$\begin{aligned}\text S = \frac{0.25 \times 55}{3 - 0.25} \\[1em] = \frac{13.75}{2.75} \\[1em] \text S = 5\ \Omega\end{aligned}$$

Hence, the galvanometer is converted into an ammeter of range 0 - 3 A by connecting a shunt of 5 Ω in parallel with its coil.

Given,

• Resistance of the coil, G = 50 Ω
• Current for full-scale deflection, I_g = 5 mA = 5 × 10^-3 A
• Required range, V = 3 V

To convert the galvanometer into a voltmeter, a resistance R is connected in series with its coil, where

$\text R = \frac{\text V}{\text I_\text g} - \text G$

Substituting the values,

$$\begin{aligned}\text R = \frac{3}{5 \times 10^{-3}} - 50 \\[1em] = 600 - 50 \\[1em] \text R = 550\ \Omega\end{aligned}$$

Hence, the galvanometer is converted into a voltmeter of range 0 - 3 V by connecting a resistance of 550 Ω in series with its coil.

Assertion is true and Reason is false.

Explanation

Assertion (A) is correct: When a current is passed through the coil, the magnetic field of the magnet exerts forces on the sides of the coil. These forces form a couple, and the deflecting torque NIAB turns the coil until it is balanced by the restoring torque of the suspension.

Reason (R) is incorrect: The magnetic field produced by a current-carrying circular coil is not uniform; it varies from point to point around the coil. Moreover, the coil is deflected by the external field of the magnet, not by its own field.

Therefore, Assertion is true and Reason is false.

Both Assertion and Reason are true and Reason is the correct explanation for Assertion.

Explanation

Assertion (A) is correct: For a moving coil galvanometer I = Kφ, so the deflection of the coil is directly proportional to the current and the scale is linear.

Reason (R) is also correct: Cylindrical pole-pieces with a coaxial cylindrical soft-iron core make the field radial. The plane of the coil then remains parallel to the field in every position, the deflecting torque is always NIAB, and NIAB = cφ gives I ∝ φ. Thus the radial field is the cause of the linear scale.

Therefore, both Assertion and Reason are true and Reason is the correct explanation for Assertion.

0.5 A

Reason — The 20 Ω shunt is in parallel with the 480 Ω coil, so the resistance of the ammeter is

$$\begin{aligned}\text R_\text A = \frac{480 \times 20}{480 + 20} \\[1em] \text R_\text A = 19.2\ \Omega\end{aligned}$$

The ammeter is in series with the 40.8 Ω resistance, so the current it reads is

$$\begin{aligned}\text I = \frac{30}{40.8 + 19.2} \\[1em] = \frac{30}{60} \\[1em] \text I = 0.5\ \text A\end{aligned}$$

$\sqrt{3}\text W$

Reason — The work done in rotating the magnet from its equilibrium position through 60° is

$$\begin{aligned}\text W = \text{MB}(1 - \cos 60^\circ) \\[1em] \text W = \frac{\text{MB}}{2} \\[1em] \text{MB} = 2\text W\end{aligned}$$

The torque needed to hold the magnet in the new position is

$$\begin{aligned}\tau = \text{MB}\sin 60^\circ \\[1em] = 2\text W \times \frac{\sqrt{3}}{2} \\[1em] \tau = \sqrt{3}\text W\end{aligned}$$

0.01 Ω

Reason — For full-scale deflection at I = 10 A, the coil carries I_g = 1 mA and the shunt S carries the rest. As G and S are in parallel, I_gG = (I − I_g)S, so

$$\begin{aligned}\text S = \frac{\text I_\text g\text G}{\text I - \text I_\text g} \\[1em] = \frac{10^{-3} \times 100}{10 - 10^{-3}} \\[1em] \text S \approx 0.01\ \Omega\end{aligned}$$

1.985 × 10³ Ω

Reason — The series resistance R required for a voltmeter of range V is

$$\begin{aligned}\text R = \frac{\text V}{\text I_\text g} - \text G \\[1em] = \frac{10}{5 \times 10^{-3}} - 15 \\[1em] = 2000 - 15 \\[1em] \text R = 1985\ \Omega \\[1em] \text R = 1.985 \times 10^3\ \Omega\end{aligned}$$

250 Ω

Reason — Current sensitivity, S_c = 5 div/mA = 5000 div/A and voltage sensitivity, S_V = 20 div/V. Since $\text S_\text V = \frac{\text S_\text c}{\text R}$,

$$\begin{aligned}\text R = \frac{\text S_\text c}{\text S_\text V} \\[1em] = \frac{5000}{20} \\[1em] \text R = 250\ \Omega\end{aligned}$$

Bπr²IN

Reason — The coil lies in the X-Z plane, so its magnetic moment M = NIπr² is along the Y-axis, while the field is along the X-axis. The angle between them is 90°, so

$$\begin{aligned}\tau = \text{MB}\sin 90^\circ \\[1em] \tau = \text B\pi\text r^2\text{IN}\end{aligned}$$

10^-3

Reason — With B parallel to the plane of the coil, the deflecting torque NIAB balances the restoring torque cφ. Here φ = 1° = $\frac{\pi}{180}$ rad, so

$$\begin{aligned}\text B = \frac{\text c\phi}{\text{NIA}} \\[1em] = \frac{10^{-6} \times \frac{3.14}{180}}{175 \times 10^{-3} \times 10^{-4}} \\[1em] = \frac{1.744 \times 10^{-8}}{1.75 \times 10^{-5}} \\[1em] \text B \approx 10^{-3}\ \text T\end{aligned}$$

0.27 Nm

Reason — In the X-Z plane, the normal to the coil is along the Y-axis, perpendicular to the field along the X-axis. On tilting the coil through 45° about the Z-axis, the normal turns in the X-Y plane and makes 45° with the field. With A = 5 cm × 2.5 cm = 12.5 × 10^-4 m²,

$$\begin{aligned}\tau = \text{NIAB}\sin 45^\circ \\[1em] = 100 \times 3 \times 12.5 \times 10^{-4} \times 1 \times 0.707 \\[1em] = 0.375 \times 0.707 \\[1em] \tau \approx 0.27\ \text{N m}\end{aligned}$$

20 V

Reason — The 5 kΩ resistance is in series with the galvanometer, so the maximum voltage that can be measured is

$$\begin{aligned}\text V = \text I_\text g(\text G + \text R) \\[1em] = 4 \times 10^{-3} \times (50 + 5000) \\[1em] = 4 \times 10^{-3} \times 5050 \\[1em] \text V = 20.2\ \text V \approx 20\ \text V\end{aligned}$$

R₁ = 1900 Ω, R₂ = 8000 Ω, R₃ = 10000 Ω

Reason — The current for full-scale deflection is I_g = 50 × 20 μA = 10^-3 A. The resistances are joined in series, and the total resistance from the galvanometer up to each tapping point must equal V/I_g for that range:

$$\begin{aligned}100 + \text R_1 = \frac{2}{10^{-3}} \\[1em] 100 + \text R_1 = 2000 \\[1em] \text R_1 = 1900\ \Omega \\[1em] 100 + 1900 + \text R_2 = \frac{10}{10^{-3}} \\[1em] 2000 + \text R_2 = 10000 \\[1em] \text R_2 = 8000\ \Omega \\[1em] 2000 + 8000 + \text R_3 = \frac{20}{10^{-3}} \\[1em] 10000 + \text R_3 = 20000 \\[1em] \text R_3 = 10000\ \Omega\end{aligned}$$

With the 2 V, 10 V and 20 V taps taken after R₁, R₂ and R₃ respectively, this is circuit (1).

$\text R_\text A \text R_\text V = \text G^2$ and $\frac{\text R_\text A}{\text R_\text V} = \left(\frac{\text I_\text g}{\text I_0 - \text I_\text g}\right)^2$

Reason — For the ammeter, $\text R_\text A = \frac{\text I_\text g\text G}{\text I_0 - \text I_\text g}$. For the voltmeter of range V = GI₀,

$$\begin{aligned}\text R_\text V = \frac{\text V}{\text I_\text g} - \text G \\[1em] = \frac{\text G\text I_0}{\text I_\text g} - \text G \\[1em] \text R_\text V = \frac{\text G(\text I_0 - \text I_\text g)}{\text I_\text g}\end{aligned}$$

Multiplying and dividing these two expressions,

$$\begin{aligned}\text R_\text A\text R_\text V = \text G^2 \\[1em] \frac{\text R_\text A}{\text R_\text V} = \left(\frac{\text I_\text g}{\text I_0 - \text I_\text g}\right)^2\end{aligned}$$

22 Ω

Reason — Let G be the resistance of the galvanometer and E the emf of the battery. The deflection is proportional to the current through G.

With K₂ open, G is in series with R₁ = 220 Ω:

$\text I_1 = \frac{\text E}{220 + \text G}$

With K₂ closed, R₂ = 5 Ω is in parallel with G, and the current through G is

$$\begin{aligned}\text I_2 = \frac{\text E}{220 + \frac{5\text G}{\text G + 5}} \times \frac{5}{\text G + 5} \\[1em] \text I_2 = \frac{5\text E}{225\text G + 1100}\end{aligned}$$

As the deflection falls to θ₀/5, I₂ = I₁/5:

$$\begin{aligned}\frac{5\text E}{225\text G + 1100} = \frac{\text E}{5(220 + \text G)} \\[1em] 25(220 + \text G) = 225\text G + 1100 \\[1em] 5500 + 25\text G = 225\text G + 1100 \\[1em] 200\text G = 4400 \\[1em] \text G = 22\ \Omega\end{aligned}$$

$\frac{2\mu_0 \text I^2 \text a^2}{\pi \text b}$

Reason — The two sides of the loop parallel to the X-axis experience forces along the Z-axis, which give no torque about the Z-axis. Each of the other two sides (length 2a, parallel to the Z-axis at x = ±a) is at a distance $\sqrt{\text a^2 + \text b^2}$ from the wire, where the field of the wire is

$\text B = \frac{\mu_0\text I}{2\pi\sqrt{\text a^2 + \text b^2}}$

The force F = BI(2a) on each of these sides lies in the X-Y plane, perpendicular to the line joining the side to the wire. Its moment arm about the Z-axis is $\frac{\text{ab}}{\sqrt{\text a^2 + \text b^2}}$, and the two torques act in the same sense. Hence

$$\begin{aligned}\tau = 2 \times \text F \times \frac{\text{ab}}{\sqrt{\text a^2 + \text b^2}} \\[1em] = 2 \times \frac{\mu_0\text I \times 2\text{aI}}{2\pi\sqrt{\text a^2 + \text b^2}} \times \frac{\text{ab}}{\sqrt{\text a^2 + \text b^2}} \\[1em] \tau = \frac{2\mu_0\text I^2\text a^2\text b}{\pi(\text a^2 + \text b^2)}\end{aligned}$$

As b >> a, a² + b² ≈ b², so

$$\begin{aligned}\tau = \frac{2\mu_0\text I^2\text a^2\text b}{\pi\text b^2} \\[1em] \tau = \frac{2\mu_0\text I^2\text a^2}{\pi\text b}\end{aligned}$$

Note: The printed key gives option 2 by retaining the exact finite-b expression, but the stated condition b >> a requires option 4.

Multimeter shows NO deflection in both cases i.e., before and after reversing the probes if the chosen component is capacitor.

Multimeter shows NO deflection in both cases i.e., before and after reversing the probes if the chosen component is metal wire.

Both options 1 and 2 describe observations that will NOT occur, so the question has no unique correct option as printed.

Reason — In the resistance mode the multimeter drives a small current through the component. A capacitor is charged by this current, so the meter shows a momentary deflection each time the probes are connected, before and after reversing them; "no deflection in both cases" is not observed. A metal wire has negligible resistance in both directions, so the meter shows a large (full-scale) deflection before and after reversing; "no deflection in both cases" is not observed here either. A resistor conducts equally both ways, and an LED conducts and glows only when forward-biased, as stated in options 3 and 4.

Note: The printed key gives option 1 only.

R₁ + G

Reason — Let R be the additional resistance. For the two ranges,

$$\begin{aligned}\text I_\text g(\text G + \text R_1) = 1 \\[1em] \text I_\text g(\text G + \text R_1 + \text R) = 2\end{aligned}$$

Dividing the second by the first,

$$\begin{aligned}\frac{\text G + \text R_1 + \text R}{\text G + \text R_1} = 2 \\[1em] \text R = \text R_1 + \text G\end{aligned}$$

3 × 10^-3 A/div.

Reason — Figure of merit is the current required per unit deflection. Taking the deflection of 2° as 2 divisions,

$$\begin{aligned}\text{Figure of merit} = \frac{\text I}{\theta} \\[1em] = \frac{6 \times 10^{-3}\ \text A}{2\ \text{div}} \\[1em] = 3 \times 10^{-3}\ \text A/\text{div}\end{aligned}$$

2 Ω

Reason — The current for full-scale deflection of 50 divisions is

$$\begin{aligned}\text I_\text g = \frac{50\ \text{div}}{2\ \text{div/mA}} \\[1em] \text I_\text g = 25\ \text{mA}\end{aligned}$$

Hence the resistance of the galvanometer is

$$\begin{aligned}\text G = \frac{\text V}{\text I_\text g} \\[1em] = \frac{50 \times 10^{-3}}{25 \times 10^{-3}} \\[1em] \text G = 2\ \Omega\end{aligned}$$

$\frac{1}{3}$ deflection method can be used and in this case, the G equals to twice the value of shunt resistance (S)

Reason — A high resistance is joined in series with the galvanometer and the deflection is noted. A shunt S is then connected across the galvanometer and adjusted until the deflection falls to one-third. As the series resistance is very large, the main current I remains practically unchanged, so the current through the galvanometer is I × S/(S + G) = I/3:

$$\begin{aligned}\frac{\text S}{\text S + \text G} = \frac{1}{3} \\[1em] 3\text S = \text S + \text G \\[1em] \text G = 2\text S\end{aligned}$$

$\frac{\text {nK}(\text G + \text S)}{\text S}$

Reason — The figure of merit K is the current per division, so for n divisions the current through the galvanometer is I_g = nK. The shunt and the galvanometer are in parallel, so I_gG = (I − I_g)S, which gives

$$\begin{aligned}\text I = \frac{\text I_\text g(\text G + \text S)}{\text S} \\[1em] \text I = \frac{\text{nK}(\text G + \text S)}{\text S}\end{aligned}$$

10%

Reason — The galvanometer and the shunt are in parallel, so the fraction of the total current I that passes through the galvanometer is

$$\begin{aligned}\frac{\text I_\text g}{\text I} = \frac{\text S}{\text S + \text G} \\[1em] = \frac{8}{8 + 72} \\[1em] = \frac{1}{10}\end{aligned}$$

that is, 10% of the total current.

Statement I is true but Statement II is false

Reason — Current sensitivity $\text S_\text c = \frac{\text{NAB}}{\text c}$, so doubling N doubles S_c; Statement I is true. Voltage sensitivity $\text S_\text V = \frac{\text{NAB}}{\text{cR}}$. Doubling the turns also doubles the length of wire and hence the resistance R of the coil, so S_V is not increased; Statement II is false.

1.0

Reason — In equilibrium the deflecting torque balances the restoring torque, NIAB = cφ. So

$$\begin{aligned}\text A = \frac{\text c\phi}{\text{NIB}} \\[1em] = \frac{4.0 \times 10^{-5} \times 0.05}{200 \times 10 \times 10^{-3} \times 0.01} \\[1em] = \frac{2.0 \times 10^{-6}}{2.0 \times 10^{-2}} \\[1em] = 10^{-4}\ \text m^2 \\[1em] \text A = 1.0\ \text{cm}^2\end{aligned}$$

$\tan \theta = \frac{\pi \text {rIB}_0}{\text {mg}}$

Reason — Let the loop turn through θ about PQ. Its magnetic moment M = Iπr² is along the normal to the loop, which now makes an angle (90° − θ) with the vertical field B₀. The weight mg acts at the centre of the loop, at a perpendicular distance r sin θ from PQ.

For equilibrium the magnetic torque balances the torque due to the weight:

$$\begin{aligned}\text{MB}_0\sin(90^\circ - \theta) = \text{mg} \times \text r\sin\theta \\[1em] \text I\pi\text r^2\text B_0\cos\theta = \text{mgr}\sin\theta \\[1em] \tan\theta = \frac{\pi\text{rIB}_0}{\text{mg}}\end{aligned}$$

Given,

• Number of turns, N = 500
• Area of each turn, A = 3 × 10^-4 m²
• Torque, τ = 1.5 N m
• Current, I = 0.5 A

The torque on a coil is τ = NIAB sin θ, where θ is the angle between the normal to the coil and the field. When the coil is parallel to the field, θ = 90°, so τ = NIAB.

$$\begin{aligned}\text B = \frac{\tau}{\text{NIA}} \\[1em] = \frac{1.5}{500 \times 0.5 \times 3 \times 10^{-4}} \\[1em] = \frac{1.5}{0.075} \\[1em] \text B = 20\ \text T\end{aligned}$$

Hence, the strength of the field is 20 T.

Given,

• Side of the triangle, a = 10 cm = 0.1 m
• Magnetic field, B = 20 mT = 20 × 10^-3 T
• Current, I = 0.2 A
• Plane of the coil parallel to the field, θ = 90°

The area of an equilateral triangle of side a is $\text A = \frac{\sqrt{3}}{4}\text a^2$. As the plane of the coil is parallel to the field, the torque is τ = IAB.

$$\begin{aligned}\tau = \text I \times \frac{\sqrt{3}}{4}\text a^2 \times \text B \\[1em] = 0.2 \times \frac{\sqrt{3}}{4} \times (0.1)^2 \times 20 \times 10^{-3} \\[1em] = \sqrt{3} \times \frac{0.2 \times 0.01 \times 0.02}{4} \\[1em] \tau = \sqrt{3} \times 10^{-5}\ \text{N m}\end{aligned}$$

Comparing with $\sqrt{\text x}$ × 10^-5 N m,

Hence, the value of x is 3.

Given,

• Total length of the wire = 24a
• Side of the triangle = side of the square = a
• The same wire (resistance R) is connected to the same source V₀ in both cases, so the current I = V₀/R is the same.

The magnetic moment of a coil is M = nIA, where n is the number of turns.

Equilateral triangle :

$$\begin{aligned}\text n_1 = \frac{24\text a}{3\text a} \\[1em] \text n_1 = 8 \\[1em] \text A_1 = \frac{\sqrt{3}}{4}\text a^2 \\[1em] \text M_1 = 8\text I \times \frac{\sqrt{3}}{4}\text a^2 \\[1em] \text M_1 = 2\sqrt{3}\thinspace\text{Ia}^2\end{aligned}$$

Square :

$$\begin{aligned}\text n_2 = \frac{24\text a}{4\text a} \\[1em] \text n_2 = 6 \\[1em] \text A_2 = \text a^2 \\[1em] \text M_2 = 6\text{Ia}^2\end{aligned}$$

Ratio :

$$\begin{aligned}\frac{\text M_1}{\text M_2} = \frac{2\sqrt{3}\thinspace\text{Ia}^2}{6\text{Ia}^2} \\[1em] = \frac{\sqrt{3}}{3} \\[1em] \frac{\text M_1}{\text M_2} = \frac{1}{\sqrt{3}}\end{aligned}$$

So M₁ : M₂ = 1 : $\sqrt{3}$.

Hence, the value of y is 3.

Given,

• Length of the wire, l = 314 cm = 3.14 m
• Current, I = 14 A
• π = 3.14

The wire forms a circle of circumference l, so its radius is

$$\begin{aligned}\text r = \frac{\text l}{2\pi} \\[1em] = \frac{3.14}{2 \times 3.14} \\[1em] \text r = 0.5\ \text m\end{aligned}$$

The magnetic moment of the coil is

$$\begin{aligned}\text M = \text{IA} \\[1em] \text M = \text I\pi\text r^2 \\[1em] = 14 \times 3.14 \times (0.5)^2 \\[1em] = 14 \times 0.785 \\[1em] = 10.99 \\[1em] \text M \approx 11\ \text A\ \text m^2\end{aligned}$$

Hence, the magnetic moment of the coil is 11 A-m².

If both assertion and reason are true and reason is the correct explanation of assertion.

Explanation

Assertion (A) is correct: In a moving coil galvanometer the coil is wound on a light, non-magnetic, metallic (such as aluminium) frame.

Reason (R) is also correct: When the coil rotates, eddy currents are induced in the metallic frame, and these oppose the motion of the coil (electromagnetic damping). The coil therefore comes to rest quickly in its deflected position without oscillating, i.e., the galvanometer becomes dead-beat. This is precisely why the frame is metallic, so the Reason explains the Assertion.

Therefore, both assertion and reason are true and reason is the correct explanation of assertion.

If both assertion and reason are true but reason is not the correct explanation of assertion.

Explanation

Assertion (A) is correct: The torque on a coil is τ = NIAB sin θ, where θ is the angle between the normal to the coil and the field. The torque is zero when θ = 0°, i.e., when the plane of the coil is perpendicular to the field, so the coil comes to equilibrium in this position.

Reason (R) is also correct: The torque is maximum (NIAB) when θ = 90°, i.e., when the plane of the coil is parallel to the field. However, this describes the position of maximum torque and does not explain the position of equilibrium, where the torque is zero.

Therefore, both assertion and reason are true but reason is not the correct explanation of assertion.

If both assertion and reason are true and reason is the correct explanation of assertion.

Explanation

Assertion (A) is correct: A rectangular coil carrying a steady current in a uniform magnetic field experiences only a torque; the net force on it is zero.

Reason (R) is also correct: The currents in opposite sides of the coil are in opposite directions, so the forces on each pair of opposite sides are equal in magnitude and opposite in direction. They cancel in pairs, which is why the net force is zero. Hence the Reason explains the Assertion.

Therefore, both assertion and reason are true and reason is the correct explanation of assertion.

If assertion is true but reason is false.

Explanation

Assertion (A) is correct: A voltmeter measures the potential difference across a part of a circuit, so it is always connected in parallel with that part.

Reason (R) is incorrect: A voltmeter draws a small current from the part of the circuit across which it is connected. The potential drop across that part therefore decreases, and the voltmeter reads slightly less, not more, than the original potential difference.

Therefore, assertion is true but reason is false.

If both assertion and reason are true but reason is not the correct explanation of assertion.

Explanation

Assertion (A) is correct: The torque on the coil is τ = NIAB sin θ. In a radial field the plane of the coil always remains parallel to the field, so θ = 90°, sin θ = 1 and the torque has its maximum value NIAB.

Reason (R) is also correct: The deflecting torque does tend to rotate the coil about its own axis of suspension. But this only states the effect of the torque; it does not explain why the torque is maximum in a radial field.

Therefore, both assertion and reason are true but reason is not the correct explanation of assertion.

If assertion is true but reason is false.

Explanation

Assertion (A) is correct: A galvanometer is a sensitive instrument used in a circuit to detect the presence of a current and its direction.

Reason (R) is incorrect: A galvanometer is not restricted to use as an ammeter. It can be converted into an ammeter by joining a low resistance in parallel with it, and into a voltmeter by joining a high resistance in series with it.

Therefore, assertion is true but reason is false.

If both assertion and reason are false.

Explanation

Assertion (A) is incorrect: A moving coil galvanometer is based on the magnetic effect of current. A current-carrying coil placed in a magnetic field experiences a torque, which deflects it.

Reason (R) is also incorrect: The coil rotates because of the couple formed by the magnetic forces on its sides, not because it is heated.

Therefore, both assertion and reason are false.

If both assertion and reason are false.

Explanation

Assertion (A) is incorrect: The magnetic moment of a circular coil is M = NIπr². When the radius becomes half, $\text M' = \text{NI}\pi\left(\frac{\text r}{2}\right)^2 = \frac{\text M}{4}$, so the magnetic moment becomes one-fourth, not four times.

Reason (R) is also incorrect: The magnetic moment M = NIA is directly proportional to the area of the loop.

Therefore, both assertion and reason are false.

If both assertion and reason are true and reason is the correct explanation of assertion.

Explanation

Assertion (A) is correct: A planar loop of area A carrying a current I behaves as a magnetic dipole of dipole moment M = IA, directed along the normal to the loop.

Reason (R) is also correct: The field of a circular loop on its axis at a distance x is $\text B = \frac{\mu_0}{4\pi}\frac{2\pi\text{Ir}^2}{(\text r^2 + \text x^2)^{3/2}}$. For x >> r this becomes $\text B = \frac{\mu_0}{4\pi}\frac{2\text M}{\text x^3}$, which is the field of a magnetic dipole of moment M = IA. This equality of fields at large distances is the basis for treating the loop as a dipole, so the Reason explains the Assertion.

Therefore, both assertion and reason are true and reason is the correct explanation of assertion.

If assertion is true but reason is false.

Explanation

Assertion (A) is correct: The shunt for range I is $\text S = \left(\frac{\text I_\text g}{\text I - \text I_\text g}\right)\text G$, which is smaller for a larger range. Hence an ammeter of higher range has a lower resistance.

Reason (R) is incorrect: To increase the range of an ammeter, an additional shunt is connected in parallel with it, not in series.

Therefore, assertion is true but reason is false.