Magnetic Field and Earth's Magnetism

a torque but no force

Reason — In a uniform magnetic field the forces on the two poles of the magnet are equal, opposite and parallel, so the net force is zero. They form a couple, which gives a torque τ = mB sin θ tending to align the magnet along the field.

magnetic declination

Reason — At any place, the acute angle between the magnetic meridian and the geographical meridian is called the angle of declination.

straight and parallel

Reason — Over a small area the earth's magnetic field is uniform, so its horizontal component is the same in magnitude and direction everywhere. Its lines of force are therefore straight, parallel and equidistant, pointing towards north.

vertical

Reason — In a vertical plane making an angle α with the magnetic meridian, the vertical component B_V is unchanged but the horizontal component becomes B_H cos α. For α = 90°,

$$\begin{aligned}\tan\theta' = \frac{\text B_\text V}{\text B_\text H\cos 90^\circ} \\[1em] \tan\theta' = \infty\end{aligned}$$

so θ' = 90° and the dip needle stays vertical.

magnetic equator

Reason — At the magnetic equator the angle of dip θ = 0°, so B_V = B_E sin 0° = 0 and the earth's field there is horizontal.

$\frac{2\text B_\text H}{\sqrt{3}}$

Reason — Since B_H = B_E cos θ, with θ = 30°,

$$\begin{aligned}\text B_\text E = \frac{\text B_\text H}{\cos 30^\circ} \\[1em] = \frac{\text B_\text H}{\left(\frac{\sqrt{3}}{2}\right)} \\[1em] \text B_\text E = \frac{2\text B_\text H}{\sqrt{3}}\end{aligned}$$

magnetic equator

Reason — At points on the magnetic equator the angle of dip is zero, so the earth's field is horizontal (parallel to the earth's surface). The circle joining all such points is the magnetic equator.

$\text B = \frac{\mu_0}{4\pi}\frac{2\text{mr}}{(\text r^2 - \text l^2)^2}$

Explanation:
Here m is the magnetic moment of the dipole, 2l is its magnetic length (distance between the poles), r is the distance of the axial point from the centre of the dipole and $\frac{\mu_0}{4\pi}$ = 10^-7 N A^-2. The field is directed along the axis, from the south pole to the north pole.

$\vec{\tau} = \vec{\text m} \times \vec{\text B}$

or, in magnitude, τ = mB sin θ.

Explanation:
Here $\vec{\text m} \space$is the magnetic moment of the magnet, $\vec{\text B} \space$is the uniform magnetic field and θ is the angle between $\vec{\text m} \space$and $\vec{\text B}$. The torque tends to align the magnet along the field.

$\text U = -\vec{\text m} \cdot \vec{\text B} = -\text{mB}\cos\theta$

The magnet is in stable equilibrium when $\vec{\text m} \space$is parallel to $\vec{\text B} \space$(θ = 0°, U = −mB) and in unstable equilibrium when $\vec{\text m} \space$is antiparallel to $\vec{\text B} \space$(θ = 180°, U = +mB).

When its magnetic moment $\vec{\text m} \space$is directed along the magnetic field $\vec{\text B} \space$(θ = 0°).

Explanation:
Since U = −mB cos θ, the potential energy has its minimum value, −mB, at θ = 0°. This is the position of stable equilibrium.

$\text B = \frac{\mu_0}{4\pi}\frac{2\text m}{\text r^3}$

where m is the magnetic moment of the bar-magnet, r is the distance of the point from the centre of the magnet and μ₀ is the permeability of free space ($\frac{\mu_0}{4\pi}$ = 10^-7 N A^-2).

$\text B = \frac{\mu_0}{4\pi}\frac{\text m}{\text r^3}$

where m is the magnetic moment of the small bar-magnet, r is the distance of the point on the equatorial line from the centre of the magnet and μ₀ is the permeability of free space ($\frac{\mu_0}{4\pi}$ = 10^-7 N A^-2). The field is parallel to the magnetic axis, directed from N to S.

No.

Explanation:
A freely-suspended magnet rests in the magnetic meridian, which makes an angle, the angle of declination, with the geographical meridian. The two do not coincide because the earth's magnetic axis is itself inclined at about 11.5° to its axis of rotation.

$\tan\theta = \frac{\text B_\text V}{\text B_\text H}$

Explanation:
Here θ is the angle of dip, and B_H and B_V are the horizontal and vertical components of the earth's magnetic field at the place, so that B_V = B_H tan θ.

Yes.

Explanation:
The angles of declination and dip change not only from place to place but also, irregularly, at the same place from time to time. The change becomes appreciable over a few hundred years.

No.

Explanation:
A magnet cannot exert a torque on itself; a torque on the magnet arises only from an external magnetic field.

$\text m = \frac{\tau}{\text B\sin\theta}$

Explanation:
The torque on the magnet is τ = mB sin θ, where m is the magnitude of its magnetic moment.

$\frac{\text B_1}{\text B_2} = \frac{\tau_1}{\tau_2}$

Explanation:
When the needle is held perpendicular to the field, θ = 90° and τ = mB sin 90° = mB. For the same needle at the two places, τ₁ = mB₁ and τ₂ = mB₂, so the fields are in the ratio of the torques.

Potential energy = −mB; work done = 2mB.

Explanation:
With $\vec{\text m} \space$along $\vec{\text B}$, θ = 0°, so

$$\begin{aligned}\text U_0 = -\text{mB}\cos 0^\circ \\[1em] \text U_0 = -\text{mB}\end{aligned}$$

After rotation through 180°, U_180° = −mB cos 180° = +mB. The work done equals the change in potential energy,

$$\begin{aligned}\text W = \text U_{180^\circ} - \text U_0 \\[1em] = \text{mB} - (-\text{mB}) \\[1em] \text W = 2\text{mB}\end{aligned}$$

The angle of dip at the magnetic equator is zero. It means that the earth's magnetic field there is horizontal, that is, its vertical component is zero.

The maximum value of the angle of dip is 90°. It occurs at the earth's magnetic poles, and it means that the earth's magnetic field at the poles is vertical (its horizontal component is zero).

45°.

Explanation:
Since tan θ = $\frac{\text B_\text V}{\text B_\text H}$ and B_V = B_H, tan θ = 1, so θ = 45°.

Horizontal component B_H = 0 and angle of dip θ = 90°.

Explanation:
At the magnetic poles the earth's field is vertical, so B_H = B_E cos 90° = 0 and B_V = B_E.

At the earth's magnetic poles.

Explanation:
A magnet pivoted to move freely in the magnetic meridian is a dip needle. It sets along the earth's field, which is vertical at the magnetic poles (angle of dip 90°).

Magnetic declination : The vertical plane passing through the axis of a magnetic needle suspended freely through its centre of gravity and at rest under the earth's field is called the magnetic meridian. The vertical plane passing through the line joining the geographical north and south poles is called the geographical meridian. The acute angle between the magnetic meridian and the geographical meridian at a place is called the angle of declination (α) at that place.

Angle of dip : The angle between the direction of the earth's magnetic field and the horizontal in the magnetic meridian at a place is called the angle of dip (θ) at that place. It is the angle which a dip needle, free to rotate in a vertical plane in the magnetic meridian, makes with the horizontal.

The tangent drawn at any point on a magnetic line of force gives the direction of the magnetic field at that point. If two lines of force crossed, there would be two tangents at the point of intersection, that is, two directions of the magnetic field at one point, which is impossible. Hence two magnetic lines of force cannot cross each other.

In a uniform magnetic field the forces on the two poles of the bar-magnet are equal, opposite and parallel. So the magnet experiences no net force, but these forces form a couple and the magnet experiences a torque (τ = mB sin θ), unless its axis is along the field.

In a non-uniform magnetic field the forces on the two poles are unequal, so the magnet experiences a net force as well as a torque.

The magnetic moment of a magnet is the quantity m in the relation τ = mB sin θ for the torque acting on the magnet in a uniform magnetic field. It is numerically equal to the torque acting on the magnet when it is held perpendicular to a uniform magnetic field of unit magnitude.

Its SI unit is ampere-metre² (A m²), which is the same as joule/tesla (J T^-1).

It is a vector quantity, $\vec{\text m}$, directed along the axis of the magnet from its south pole to its north pole.

The plane perpendicular to the magnetic axis of the earth and passing through the centre of the earth intersects the earth's spherical surface in a circle. This circle is called the geomagnetic (magnetic) equator. At every point on it the magnetic needle is parallel to the earth's surface, so the angle of dip there is zero.

The three magnetic elements are :

(i) the angle of declination (α),

(ii) the angle of dip (θ),

(iii) the horizontal component of the earth's magnetic field (B_H).

α and θ fix the direction of the earth's field B_E, and B_H together with θ gives its magnitude, B_E = $\frac{\text B_\text H}{\cos\theta}$.

Place one bar, say A, on the table. Bring one end of bar B near one end of A, and then near the other end of A.

• If B experiences attraction in both cases, only one of the two bars is magnetised; the other is not.

• If B experiences repulsion in any one case, both A and B are magnetised, since repulsion occurs only between like poles of two magnets.

Place one bar, say A, on the table and bring one end of bar B to the middle of bar A.

• If B experiences no force, then B is unmagnetised and A is magnetised. The middle of a magnet has practically no magnetic strength, so it cannot attract the unmagnetised bar B.

• If B experiences attraction, then B is magnetised and A is unmagnetised. The pole of B attracts the iron bar A at any point, including its middle.

In a uniform magnetic field the two poles of the magnetised needle experience equal, opposite and parallel forces. These forces give no resultant force, but they form a couple and produce a torque.

The field of a bar-magnet is non-uniform. The iron nail placed near it acquires induced magnetism, and the forces on its two induced poles are unequal. The attractive force on the induced pole nearer the bar-magnet, where the field is stronger, exceeds the repulsive force on the farther pole. The two unequal forces therefore give a resultant force of attraction as well as a torque.

It will be more in England.

As we go from the magnetic equator to a magnetic pole, the angle of dip increases from 0° to 90°. England is much closer to the magnetic north pole of the earth than South India, so the angle of dip there is larger (about 70°).

It can point in any direction.

A compass needle moves in a horizontal plane under the influence of the horizontal component B_H of the earth's magnetic field. At the geomagnetic poles the earth's field is vertical, so B_H = 0. No horizontal field acts on the needle, and it can stay in any direction.

The earth's magnetic field is vertical (perpendicular to the earth's surface) at the magnetic poles and horizontal (parallel to the earth's surface) at the magnetic equator.

The magnetic force on a charged particle is F = qvB sin θ. Particles arriving at the poles move nearly along the field (θ ≈ 0°), so F ≈ 0 and they are not deflected. Particles arriving at the equator move perpendicular to the field (θ = 90°), so they experience the maximum force and are deflected sideways. Hence only high-energy particles reach the equator, whereas even low-energy particles reach the poles.

(i) It is a south magnetic pole. The north pole of a freely suspended magnetic needle points towards the earth's northern hemisphere, so the earth's magnet must have its south pole there.

(ii) The lines of force point towards the earth's surface. Magnetic lines of force enter a magnet at its south pole, and the earth's magnetic south pole lies in the northern hemisphere.

The lines would seem to come out of the earth.

Melbourne is in the southern hemisphere, where the north pole of the earth's magnet lies. Magnetic lines of force emerge from a north pole, so in the southern hemisphere they come out of the earth's surface.

Close to the earth, the earth's field is very nearly a dipole field. At large distances it is modified by the magnetic field produced by the motion of ions in the earth's ionosphere. This field is sensitive to extra-terrestrial disturbances such as the solar wind, so beyond about 30000 km the earth's field departs from its dipole shape.

The earth's magnetic field is only approximately a dipole field. Magnetised mineral deposits within the earth's crust produce their own magnetic fields, which give rise to local N-S poles oriented in different directions.

Yes. A charged particle moving in a magnetic field B follows a circular arc of radius

$\text r = \frac{\text{mv}}{\text{qB}}$

In such a weak field this radius is extremely large, so the bending is not noticeable over a small distance. Over the gigantic interstellar distances, however, the deflection becomes significant and can appreciably affect the passage of charged particles such as cosmic rays.

Given,

• Magnetic moment of the dipole, m = 8 × 10²² J T^-1 = 8 × 10²² A m²
• Radius of the earth, r = 6.4 × 10⁶ m
• $\frac{\mu_0}{4\pi}$ = 10^-7 N A^-2

A point on the magnetic equator of the earth's surface is in the broad-side-on position of the dipole at its centre. The field there is

$\text B = \frac{\mu_0}{4\pi}\frac{\text m}{\text r^3}$

Substituting the values,

$$\begin{aligned}\text B = 10^{-7} \times \frac{8 \times 10^{22}}{(6.4 \times 10^6)^3} \\[1em] = 10^{-7} \times \frac{8 \times 10^{22}}{2.62 \times 10^{20}} \\[1em] = 3.05 \times 10^{-5}\ \text T \\[1em] \text B \approx 0.3 \times 10^{-4}\ \text T \\[1em] \text B \approx 0.3\ \text G\end{aligned}$$

Hence, the field comes out to about 0.3 G, which is the actual order of the earth's magnetic field. So the dipole moment 8 × 10²² J T^-1 is of the correct order of magnitude.

The earth behaves as if a powerful magnet were placed within it, with its south pole S towards the north of the earth and its north pole N towards the south.

For a current-loop to produce this field :

Plane of the loop : The axis of the loop must lie along the earth's magnetic axis, so the loop lies in the east-west plane, that is, nearly in the equatorial plane of the earth.

Direction of current : Viewed from the north, the current must flow clockwise (from east to west). Then, by the end rule, the northern face of the loop behaves as a south pole and the southern face as a north pole, as the earth's field requires.

The exact source is not known. One possibility is the radioactivity in the interior of the earth. Radioactive materials produce charged particles, which rotate with the earth and so constitute currents that can sustain the earth's magnetism.

The earth's magnetic field gets weakly recorded in certain rocks during their solidification, the rocks becoming magnetised along the field prevailing at that time. Analysis of the magnetism of such rocks of different ages reveals the direction of the earth's field in the distant past. Rocks showing magnetisation in opposite directions indicate that the earth's field has reversed its direction several times.

Consider a bar-magnet placed in a uniform magnetic field $\vec{\text B}$, with its magnetic axis making an angle θ with the direction of $\vec{\text B}$.

Torque : According to the modern view, each atom of the magnet is a tiny current-loop, and all these loops are aligned in the same direction, with their axes along the magnetic axis. The torque on one such loop of area A carrying current I is

$\tau_1 = \text{IAB}\sin\theta$

If the magnet contains N such current-loops, the torques on all of them act in the same sense and add up. So the torque on the whole magnet is

$$\begin{aligned}\tau = \text N \times \text{IAB}\sin\theta \\[1em] \tau = (\text{NIA})\thinspace\text B\sin\theta\end{aligned}$$

The quantity NIA is defined as the magnitude of the magnetic dipole moment m of the magnet. Thus,

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

or, in vector form, $\vec{\tau} = \vec{\text m} \times \vec{\text B}$. This torque tends to rotate the magnet so that its axis becomes parallel to the field.

Magnetic moment : Putting θ = 90° and B = 1 in Eq. (i) gives m = τ. Hence, the magnetic moment of a magnet is numerically equal to the torque acting on the magnet when it is held perpendicular to a uniform magnetic field of unit magnitude. It is a vector quantity directed from the south pole to the north pole of the magnet.

Unit : From Eq. (i),

$\text m = \frac{\tau}{\text B\sin\theta}$

so the SI unit of magnetic moment is $\frac{\text{N m}}{\text{N A}^{-1}\ \text m^{-1}}$ = A m², which is also written as J T^-1.

Consider a short bar-magnet NS of magnetic length 2l, with poles of strength +q_m (at N) and −q_m (at S), so that its magnetic moment is m = q_m × 2l. The magnetic field due to a pole of strength q_m at a distance x from it is $\frac{\mu_0}{4\pi}\frac{\text q_\text m}{\text x^2}$, directed away from a north pole and towards a south pole.

(i) Point on the axis (end-on position) :

Let P be a point on the axis of the magnet, at a distance r from its centre O, on the side of the north pole.

The distance of P from N is (r − l) and from S is (r + l). The field at P due to N is directed away from N, and that due to S is directed towards S, that is, the two fields are opposite to each other along the axis. Since N is nearer, the resultant field is

$$\begin{aligned}\text B = \frac{\mu_0}{4\pi}\frac{\text q_\text m}{(\text r - \text l)^2} - \frac{\mu_0}{4\pi}\frac{\text q_\text m}{(\text r + \text l)^2} \\[1em] = \frac{\mu_0\text q_\text m}{4\pi}\left[\frac{(\text r + \text l)^2 - (\text r - \text l)^2}{(\text r^2 - \text l^2)^2}\right] \\[1em] = \frac{\mu_0\text q_\text m}{4\pi}\frac{4\text{rl}}{(\text r^2 - \text l^2)^2} \\[1em] \text B = \frac{\mu_0}{4\pi}\frac{2(\text q_\text m \times 2\text l)\text r}{(\text r^2 - \text l^2)^2}\end{aligned}$$

Putting q_m × 2l = m,

$\text B = \frac{\mu_0}{4\pi}\frac{2\text{mr}}{(\text r^2 - \text l^2)^2}$

For a short magnet (l << r), l² may be neglected in comparison with r², so

$\text B_\text{axial} = \frac{\mu_0}{4\pi}\frac{2\text m}{\text r^3} \qquad ...(\text i)$

The field is along the axis, directed from S to N (along $\vec{\text m}$).

(ii) Point on the equatorial line (broad-side-on position) :

Let P be a point on the equatorial line, at a distance r from the centre O. Each pole is at a distance $\sqrt{\text r^2 + \text l^2}$ from P.

The magnitudes of the fields at P due to the two poles are equal,

$\text B_\text N = \text B_\text S = \frac{\mu_0}{4\pi}\frac{\text q_\text m}{\text r^2 + \text l^2}$

B_N is directed away from N along NP, and B_S towards S along PS. Let each make an angle φ with the axis. Their components perpendicular to the axis are equal and opposite and cancel, while their components parallel to the axis add up, both pointing from N to S. Hence

$$\begin{aligned}\text B = 2\text B_\text N\cos\varphi \\[1em] = 2 \times \frac{\mu_0}{4\pi}\frac{\text q_\text m}{\text r^2 + \text l^2} \times \frac{\text l}{\sqrt{\text r^2 + \text l^2}} \\[1em] = \frac{\mu_0}{4\pi}\frac{\text q_\text m \times 2\text l}{(\text r^2 + \text l^2)^{3/2}} \\[1em] \text B = \frac{\mu_0}{4\pi}\frac{\text m}{(\text r^2 + \text l^2)^{3/2}}\end{aligned}$$

For a short magnet (l << r),

$\text B_\text{equatorial} = \frac{\mu_0}{4\pi}\frac{\text m}{\text r^3} \qquad ...(\text{ii})$

The field is parallel to the axis, directed from N to S (opposite to $\vec{\text m}$).

Hence, for a short bar-magnet, the field at an axial point is $\frac{\mu_0}{4\pi}\frac{2\text m}{\text r^3}$ and at an equatorial point is $\frac{\mu_0}{4\pi}\frac{\text m}{\text r^3}$; at equal distances the axial field is twice the equatorial field.

Elements : The three elements of the earth's magnetic field are :

(i) Angle of declination (α) — the acute angle between the magnetic meridian and the geographical meridian at a place.

(ii) Angle of dip (θ) — the angle between the direction of the earth's magnetic field and the horizontal in the magnetic meridian at a place.

(iii) Horizontal component (B_H) — the component of the earth's magnetic field in the horizontal direction in the magnetic meridian.

In the diagram, OPQR is the magnetic meridian and OLMR is the geographical meridian; the angle between them is α. OQ is the direction of the earth's field B_E, making the angle θ with the horizontal OP. B_E resolves into B_H along OP and B_V vertically downwards.

Importance : If α and θ are known at a place, the direction of the earth's field B_E is fixed; if B_H and θ are known, its magnitude can be found. Thus α, θ and B_H together give full information about the earth's magnetic field at that place.

Relations :

$$\begin{aligned}\text B_\text H = \text B_\text E\cos\theta \\[1em] \text B_\text V = \text B_\text E\sin\theta \\[1em] \text B_\text E = \sqrt{\text B_\text H^2 + \text B_\text V^2} \\[1em] \tan\theta = \frac{\text B_\text V}{\text B_\text H}\end{aligned}$$

Variation of dip : The angle of dip is zero at the magnetic equator, where the field is horizontal. It increases as we move towards a magnetic pole and becomes 90° at the magnetic poles, where the field is vertical.

Magnetic lines of force : A magnetic line of force is an imaginary curve in a magnetic field along which the direction of the field is represented continuously, so that the tangent at any point on it gives the direction of the magnetic field at that point.

Properties :

(i) Outside the magnet they emerge from the north pole and curve round into the south pole; inside the magnet they continue from the south pole back to the north pole. Each line is therefore a closed loop, whereas electric lines of force are not closed loops.

(ii) Lines of force do not intersect: a crossing would mean two tangents, and hence two directions of the magnetic field, at the same point.

(iii) The crowding of the lines indicates the strength of the field. They are close together near the poles, where the field is strong, and become rarer farther away, where it is weaker; a greater number of lines of force at a point means a stronger magnetic field there.

(iv) A uniform magnetic field is represented by parallel and equidistant lines of force.

(v) A neutral point has no line of force passing through it.

(vi) The angle at which a line of force enters or leaves a magnetic pole is not fixed; it can be any angle.

(vii) Lines of force running between opposite (N-S) poles behave like stretched strings trying to contract, which accounts for attraction; lines between similar (N-N or S-S) poles tend to repel one another, which accounts for repulsion.

Tracing the lines for a bar-magnet : Place the bar-magnet on a sheet of paper and a small compass needle near its north pole. Mark the positions of the two ends of the needle with dots. Shift the needle so that its south end lies at the dot marking its earlier north end, and mark the new position of the north end. Repeat this step until the needle reaches the south pole of the magnet. The smooth curve joining the dots is one line of force, its direction being from N to S outside the magnet. Repeat from different starting points near the north pole to trace more lines.

The lines of force of a bar-magnet emerge from its north pole and curve round to enter its south pole, as shown.

In terms of current and area of a loop : A current-loop of area A carrying a current I behaves as a magnetic dipole whose magnetic dipole moment is

$\vec{\text m} = \text I\vec{\text A}$

that is, the magnetic dipole moment of a current-loop is the product of the current in the loop and the area of the loop. Its direction is perpendicular to the plane of the loop, given by the right-hand rule : if the fingers of the right hand curl around the loop in the direction of the current, the extended thumb points along $\vec{\text m}$. For a coil of N turns, $\vec{\text m} \space$= NI$\vec{\text A}$.

In terms of pole strength : A magnetic dipole (bar-magnet) has two equal and opposite poles of strength q_m separated by its magnetic length 2l. Its magnetic dipole moment is

$\text m = \text q_\text m \times 2\text l$

that is, the magnetic dipole moment is the product of the pole strength and the magnetic length. It is directed along the axis from the south pole to the north pole.

Units :

• SI unit of magnetic dipole moment = ampere-metre² (A m²), also written as J T^-1.

• SI unit of pole strength = $\frac{\text{A m}^2}{\text m}$ = ampere-metre (A m).

Comparison of a current-carrying loop and a bar-magnet as magnetic dipoles :

Thus both behave identically as magnetic dipoles; according to the modern view, a bar-magnet is itself an assembly of tiny aligned current-loops.

Magnetic field on the equatorial line :

Consider a magnetic dipole NS of magnetic length 2l with poles of strength +q_m and −q_m, so that m = q_m × 2l. Let P be a point on its equatorial line at a distance r from the centre O. The field due to a pole of strength q_m at a distance x from it is $\frac{\mu_0}{4\pi}\frac{\text q_\text m}{\text x^2}$.

Each pole is at a distance $\sqrt{\text r^2 + \text l^2}$ from P, so the fields at P due to the two poles have equal magnitudes,

$\text B_\text N = \text B_\text S = \frac{\mu_0}{4\pi}\frac{\text q_\text m}{\text r^2 + \text l^2} \qquad ...(\text i)$

B_N is directed away from N along NP, and B_S towards S along PS. Let each make an angle φ with the axis of the dipole, where

$\cos\varphi = \frac{\text l}{\sqrt{\text r^2 + \text l^2}} \qquad ...(\text{ii})$

Resolving the two fields, the components B_N sin φ and B_S sin φ perpendicular to the axis are equal and opposite and cancel. The components B_N cos φ and B_S cos φ parallel to the axis are in the same direction (from N to S) and add. Hence the resultant field at P is

$$\begin{aligned}\text B = \text B_\text N\cos\varphi + \text B_\text S\cos\varphi \\[1em] \text B = 2\text B_\text N\cos\varphi\end{aligned}$$

Using Eqs. (i) and (ii),

$$\begin{aligned}\text B = 2 \times \frac{\mu_0}{4\pi}\frac{\text q_\text m}{\text r^2 + \text l^2} \times \frac{\text l}{\sqrt{\text r^2 + \text l^2}} \\[1em] = \frac{\mu_0}{4\pi}\frac{\text q_\text m \times 2\text l}{(\text r^2 + \text l^2)^{3/2}} \\[1em] \text B = \frac{\mu_0}{4\pi}\frac{\text m}{(\text r^2 + \text l^2)^{3/2}}\end{aligned}$$

For a short dipole (l << r), l² may be neglected in comparison with r², so

$\text B = \frac{\mu_0}{4\pi}\frac{\text m}{\text r^3}$

Hence, the magnetic field at a point on the equatorial line is $\frac{\mu_0}{4\pi}\frac{\text m}{(\text r^2 + \text l^2)^{3/2}}$ (reducing to $\frac{\mu_0}{4\pi}\frac{\text m}{\text r^3}$ for a short dipole), directed parallel to the axis from N to S.

At a place, the earth's total magnetic field B_E acts in the magnetic meridian, along the axis of a freely pivoted dip needle, making an angle θ with the horizontal. It is resolved into :

• the horizontal component B_H, along the horizontal line in the magnetic meridian (pointing towards magnetic north),

• the vertical component B_V, acting vertically (downwards in the northern hemisphere).

The angle θ between B_E and B_H is the angle of dip. From the diagram,

$$\begin{aligned}\text B_\text H = \text B_\text E\cos\theta \\[1em] \text B_\text V = \text B_\text E\sin\theta \\[1em] \tan\theta = \frac{\text B_\text V}{\text B_\text H}\end{aligned}$$

(i)

(a)

• The horizontal component B_H is maximum at the magnetic equator (where B_H = B_E) and minimum (zero) at the magnetic poles.

• The vertical component B_V is maximum at the magnetic poles (where B_V = B_E) and minimum (zero) at the magnetic equator.

(b)

• The angle of dip is maximum (90°) at the magnetic poles, where the earth's field is vertical.

• The angle of dip is minimum (zero) at the magnetic equator, where the earth's field is horizontal.

These follow from B_H = B_E cos θ and B_V = B_E sin θ, with θ = 0° at the magnetic equator and θ = 90° at the magnetic poles.

(ii) The exact cause of the earth's magnetism is not yet known. The earth behaves as if a powerful magnet were placed within it, with its south pole towards the earth's north and its north pole towards the earth's south. This cannot be a permanent iron magnet, because the temperature of the earth's core is above the Curie temperature of iron, at which iron cannot retain its magnetism. The earth's magnetism is thought to be due to charged currents in the outer conducting regions of the earth's core. One possible source of energy sustaining these currents is radioactivity in the interior of the earth : the charged particles so produced rotate with the earth and constitute currents.

Angle of dip : The angle between the direction of the earth's magnetic field and the horizontal in the magnetic meridian at a place is called the angle of dip (θ) at that place.

Magnetic meridian : The vertical plane passing through the axis of a magnetic needle suspended freely through its centre of gravity and at rest under the earth's field is called the magnetic meridian.

Geographic meridian : The vertical plane passing through the line joining the geographical north and south poles of the earth is called the geographic meridian.

Angle of declination : The acute angle between the magnetic meridian and the geographic meridian at a place is called the angle of declination (α) at that place.

Relation tan θ = $\frac{\text B_\text V}{\text B_\text H}$ :

Let a dip needle ns be pivoted at O so that it is free to rotate in the vertical plane OPQR, the magnetic meridian. The axis OQ of the needle, in its steady state, gives the direction of the earth's total magnetic field B_E, and the angle between OQ and the horizontal OP is the angle of dip θ.

Resolving B_E into a horizontal component B_H along OP and a vertical component B_V along OR,

$$\begin{aligned}\text B_\text H = \text B_\text E\cos\theta \qquad ...(\text i) \\[1em] \text B_\text V = \text B_\text E\sin\theta \qquad ...(\text{ii})\end{aligned}$$

Dividing Eq. (ii) by Eq. (i),

$$\begin{aligned}\frac{\text B_\text V}{\text B_\text H} = \frac{\text B_\text E\sin\theta}{\text B_\text E\cos\theta} \\[1em] \frac{\text B_\text V}{\text B_\text H} = \tan\theta\end{aligned}$$

Hence, tan θ = $\frac{\text B_\text V}{\text B_\text H}$, where θ is the angle of dip.

Given,

• Magnetic moment of the needle, m = 4.8 × 10^-2 J T^-1

• Magnetic field, B = 3.0 × 10^-2 T

• Angle between $\vec{\text m} \space$and $\vec{\text B}$, θ = 30°

The magnitude of the torque acting on a magnetic dipole placed in a uniform magnetic field is

$\tau = \text m\text B\sin\theta$

Substituting the given values,

$$\begin{aligned}\tau = 4.8 \times 10^{-2} \times 3.0 \times 10^{-2} \times \sin 30^\circ \\[1em] = 14.4 \times 10^{-4} \times 0.5 \\[1em] \tau = 7.2 \times 10^{-4}\ \text J\end{aligned}$$

In vector form $\vec{\tau} = \vec{\text m} \times \vec{\text B}$, so the torque acts perpendicular to the plane containing $\vec{\text m} \space$and $\vec{\text B}$. It tends to rotate the needle so as to bring its magnetic moment along $\vec{\text B}$.

Hence, the torque on the needle is 7.2 × 10^-4 J, directed perpendicular to the plane of $\vec{\text m} \space$and $\vec{\text B} \space$such that it tends to bring the needle along $\vec{\text B}$.

Given,

• Magnetic moment, m = 0.9 J T^-1

• Angle between the axis and the field, θ = 45°

• Torque, τ = 0.063 J

Strength of the magnetic field :

The torque acting on the magnet is

$\tau = \text m\text B\sin\theta$

so that

$$\begin{aligned}\text B = \frac{\tau}{\text m\sin\theta} \\[1em] = \frac{0.063}{0.9 \times \sin 45^\circ} \\[1em] = \frac{0.063}{0.9 \times 0.7071} \\[1em] = \frac{0.063}{0.6364} \\[1em] \text B = 0.099\ \text T\end{aligned}$$

Orientation for stable equilibrium :

The potential energy of the magnet is U = −mB cos θ. It is minimum (U = −mB) at θ = 0°, where the torque mB sin θ is also zero. Hence the magnet is in stable equilibrium when its magnetic moment $\vec{\text m} \space$is parallel to $\vec{\text B}$.

Hence, the strength of the magnetic field is 0.099 T, and the orientation of zero torque with the magnet parallel to the field (θ = 0°) corresponds to stable equilibrium.

Given,

• Magnetic field, B = 0.3 T

• Angle between the axis and the field, θ = 30°

• Torque, τ = 0.06 N m

The torque acting on a bar-magnet in a uniform magnetic field is

$\tau = \text m\text B\sin\theta$

Therefore,

$$\begin{aligned}\text m = \frac{\tau}{\text B\sin\theta} \\[1em] = \frac{0.06}{0.3 \times \sin 30^\circ} \\[1em] = \frac{0.06}{0.3 \times 0.5} \\[1em] = \frac{0.06}{0.15} \\[1em] \text m = 0.4\ \text{A m}^2\end{aligned}$$

Hence, the magnetic moment of the bar-magnet is 0.4 A m².

Given,

• Magnetic moment, m = 0.32 J T^-1

• Magnetic field, B = 0.15 T

The potential energy of a magnet whose moment $\vec{\text m} \space$makes an angle θ with $\vec{\text B} \space$is

$\text U = -\vec{\text m} \cdot \vec{\text B} = -\text m\text B\cos\theta$

In both the orientations below the torque mB sin θ is zero, so the magnet is in equilibrium.

(i) Stable equilibrium :

The potential energy is minimum when $\vec{\text m} \space$is parallel to $\vec{\text B} \space$(θ = 0°). This is the stable equilibrium.

$$\begin{aligned}\text U_0 = -\text m\text B\cos 0^\circ \\[1em] = -0.32 \times 0.15 \times 1 \\[1em] \text U_0 = -0.048\ \text J\end{aligned}$$

(ii) Unstable equilibrium :

The potential energy is maximum when $\vec{\text m} \space$is anti-parallel to $\vec{\text B} \space$(θ = 180°). This is the unstable equilibrium.

$$\begin{aligned}\text U_{180^\circ} = -\text m\text B\cos 180^\circ \\[1em] = -0.32 \times 0.15 \times (-1) \\[1em] \text U_{180^\circ} = +0.048\ \text J\end{aligned}$$

Hence, (i) for $\vec{\text m} \space$parallel to $\vec{\text B} \space$(stable equilibrium) U = −mB = −0.048 J, and (ii) for $\vec{\text m} \space$anti-parallel to $\vec{\text B} \space$(unstable equilibrium) U = +mB = +0.048 J.

Given,

• Magnetic moment, m = 1.5 A m²

• Magnet perpendicular to the field, θ = 90°

• Couple (torque), τ = 0.24 N-m

• Number of turns of the coil, N = 100

• Area of the coil, A = 150 cm² = 150 × 10^-4 m²

(a) Magnitude of the magnetic field :

The torque on the magnet is τ = mB sin θ. With θ = 90°,

$$\begin{aligned}\tau = \text m\text B\sin 90^\circ \\[1em] \tau = \text m\text B\end{aligned}$$

so that

$$\begin{aligned}\text B = \frac{\tau}{\text m} \\[1em] = \frac{0.24}{1.5} \\[1em] \text B = 0.16\ \text T\end{aligned}$$

(b) Change in the potential energy :

The potential energy of the magnet is U = −mB cos θ. In the given position θ = 90°, and in the position of steady (stable) equilibrium θ = 0°.

$$\begin{aligned}\text U_{90^\circ} = -\text m\text B\cos 90^\circ \\[1em] \text U_{90^\circ} = 0 \\[1em] \text U_0 = -\text m\text B\cos 0^\circ \\[1em] \text U_0 = -\text m\text B\end{aligned}$$

Therefore, the change in the potential energy is

$$\begin{aligned}\Delta\text U = \text U_0 - \text U_{90^\circ} \\[1em] = -\text m\text B - 0 \\[1em] = -1.5 \times 0.16 \\[1em] \Delta\text U = -0.24\ \text J\end{aligned}$$

The negative sign shows that the potential energy decreases.

(c) Current in the equivalent coil :

The magnetic moment of a current-carrying coil of N turns and area A is m = NIA. Therefore,

$$\begin{aligned}\text I = \frac{\text m}{\text{NA}} \\[1em] = \frac{1.5}{100 \times 150 \times 10^{-4}} \\[1em] = \frac{1.5}{1.5} \\[1em] \text I = 1.0\ \text A\end{aligned}$$

Hence, (a) the magnetic field is 0.16 T, (b) the potential energy decreases by 0.24 J, and (c) the current in the coil is 1.0 A.

Given,

• Magnetic moment, m = 4.0 A m²

• Distance of the axial point, r₁ = 5.0 cm = 0.05 m

• Field at the equatorial point, B₂ = 0.05 T

• $\frac{\mu_0}{4\pi}$ = 10^-7 N A^-2

Field on the axis :

For a short bar-magnet, the magnetic field at an axial (end-on) point distant r from its centre is

$\text B = \frac{\mu_0}{4\pi}\frac{2\text m}{\text r^3}$

Substituting the values,

$$\begin{aligned}\text B_1 = 10^{-7} \times \frac{2 \times 4.0}{(0.05)^3} \\[1em] = \frac{8 \times 10^{-7}}{1.25 \times 10^{-4}} \\[1em] \text B_1 = 6.4 \times 10^{-3}\ \text T\end{aligned}$$

Distance of the equatorial point :

At an equatorial (broad-side-on) point distant r from the centre,

$\text B = \frac{\mu_0}{4\pi}\frac{\text m}{\text r^3}$

Therefore,

$$\begin{aligned}\text r_2^3 = \frac{\mu_0}{4\pi}\frac{\text m}{\text B_2} \\[1em] = \frac{10^{-7} \times 4.0}{0.05} \\[1em] = 8 \times 10^{-6}\ \text m^3 \\[1em] \text r_2 = 2 \times 10^{-2}\ \text m \\[1em] \text r_2 = 2.0\ \text{cm}\end{aligned}$$

Hence, the magnetic field on the axis is 6.4 × 10^-3 T, and the point on the equatorial line is 2.0 cm from the centre of the magnet.

Given,

• Magnetic moments, m₁ = 0.20 J T^-1 and m₂ = 0.40 J T^-1

• Distance between the magnets = 20 cm

• Distance of the middle-point from each magnet, r = 10 cm = 0.10 m

The two magnets lie along the same straight line, so the middle-point P lies on the axis of each magnet. For a short magnet, the field at an axial point is

$\text B = \frac{\mu_0}{4\pi}\frac{2\text m}{\text r^3}$

and it is directed along the axis from the south pole to the north pole, that is, away from the north pole on the side of the north pole.

Field due to each magnet :

$$\begin{aligned}\text B_1 = 10^{-7} \times \frac{2 \times 0.20}{(0.10)^3} \\[1em] = \frac{0.4 \times 10^{-7}}{10^{-3}} \\[1em] \text B_1 = 4.0 \times 10^{-5}\ \text T\end{aligned}$$

$$\begin{aligned}\text B_2 = 10^{-7} \times \frac{2 \times 0.40}{(0.10)^3} \\[1em] = \frac{0.8 \times 10^{-7}}{10^{-3}} \\[1em] \text B_2 = 8.0 \times 10^{-5}\ \text T\end{aligned}$$

Resultant field :

As the north poles face each other, B₁ at P points away from the first magnet (towards the second) and B₂ points away from the second magnet (towards the first). The two fields are opposite, so

$$\begin{aligned}\text B = \text B_2 - \text B_1 \\[1em] = 8.0 \times 10^{-5} - 4.0 \times 10^{-5} \\[1em] \text B = 4.0 \times 10^{-5}\ \text T\end{aligned}$$

Hence, the magnetic field at the middle-point is 4.0 × 10^-5 T, directed along the line towards the magnet of moment 0.20 J T^-1.

Given,

• Magnetic moments, m₁ = 0.108 J T^-1 and m₂ = 0.192 J T^-1

• Distances of the point of intersection O, r₁ = 30 cm = 0.30 m and r₂ = 40 cm = 0.40 m

Each magnet lies with its axis along its own line, so the point of intersection O is an axial (end-on) point of both the magnets.

For a short magnet, the field at an axial point is

$\text B = \frac{\mu_0}{4\pi}\frac{2\text m}{\text r^3}$

Field due to each magnet :

$$\begin{aligned}\text B_1 = 10^{-7} \times \frac{2 \times 0.108}{(0.30)^3} \\[1em] = \frac{0.216 \times 10^{-7}}{0.027} \\[1em] \text B_1 = 8.0 \times 10^{-7}\ \text T\end{aligned}$$

$$\begin{aligned}\text B_2 = 10^{-7} \times \frac{2 \times 0.192}{(0.40)^3} \\[1em] = \frac{0.384 \times 10^{-7}}{0.064} \\[1em] \text B_2 = 6.0 \times 10^{-7}\ \text T\end{aligned}$$

Resultant field :

B₁ is along the first line and B₂ along the second, so the two fields are mutually perpendicular. The resultant field is

$$\begin{aligned}\text B = \sqrt{\text B_1^2 + \text B_2^2} \\[1em] = \sqrt{(8.0 \times 10^{-7})^2 + (6.0 \times 10^{-7})^2} \\[1em] = 10^{-7}\sqrt{64 + 36} \\[1em] = 10 \times 10^{-7} \\[1em] \text B = 1.0 \times 10^{-6}\ \text T\end{aligned}$$

The question does not state which way the poles of the magnets point, so only the magnitude of the resultant field is fixed; its direction depends on the orientations of the two magnets.

Hence, the resultant magnetic field at the point of intersection is 1.0 × 10^-6 T.

Given,

• Horizontal component, B_H = $\sqrt{3}$ B_V

If θ is the angle of dip, then

$\tan\theta = \frac{\text B_\text V}{\text B_\text H}$

Substituting B_H = $\sqrt{3}$ B_V,

$$\begin{aligned}\tan\theta = \frac{\text B_\text V}{\sqrt{3}\thinspace\text B_\text V} \\[1em] \tan\theta = \frac{1}{\sqrt{3}} \\[1em] \theta = 30^\circ\end{aligned}$$

Hence, the angle of dip at that place is 30°.

Given,

• $\frac{\text B_\text H}{\text B_\text E} = \frac{1}{\sqrt{2}}$

If θ is the angle of dip, the horizontal component of earth's magnetic field is

$\text B_\text H = \text B_\text E\cos\theta$

Therefore,

$$\begin{aligned}\cos\theta = \frac{\text B_\text H}{\text B_\text E} \\[1em] \cos\theta = \frac{1}{\sqrt{2}} \\[1em] \theta = 45^\circ\end{aligned}$$

Hence, the angle of dip at that place is 45°.

Given,

• Angle of dip, θ = 60°

• Horizontal component, B_H = 0.3 × 10^-4 Wb/m²

The needle rotates in the magnetic meridian, so the angle its axis makes with the horizontal is the angle of dip.

Vertical component :

$\tan\theta = \frac{\text B_\text V}{\text B_\text H}$

Therefore,

$$\begin{aligned}\text B_\text V = \text B_\text H\tan\theta \\[1em] = 0.3 \times 10^{-4} \times \tan 60^\circ \\[1em] = 0.3 \times 10^{-4} \times 1.732 \\[1em] \text B_\text V = 0.52 \times 10^{-4}\ \text{Wb/m}^2\end{aligned}$$

Total magnetic field :

$\text B_\text H = \text B_\text E\cos\theta$

Therefore,

$$\begin{aligned}\text B_\text E = \frac{\text B_\text H}{\cos\theta} \\[1em] = \frac{0.3 \times 10^{-4}}{\cos 60^\circ} \\[1em] = \frac{0.3 \times 10^{-4}}{0.5} \\[1em] \text B_\text E = 0.60 \times 10^{-4}\ \text{Wb/m}^2\end{aligned}$$

Hence, the vertical component of earth's magnetic field is 0.52 × 10^-4 Wb/m² and the total magnetic field is 0.60 × 10^-4 Wb/m².

Given,

• Angle of declination, α = 3.5° (west of geographic north)

• Angle of dip, θ = 18°

• Horizontal component, B_H = 0.40 G

• tan 18° = 0.325

The compass sets itself in the magnetic meridian, so the magnetic meridian is 3.5° west of the geographic meridian. The dip needle, rotating in the magnetic meridian, gives the angle of dip θ = 18°.

Vertical component :

$\tan\theta = \frac{\text B_\text V}{\text B_\text H}$

Therefore,

$$\begin{aligned}\text B_\text V = \text B_\text H\tan\theta \\[1em] = 0.40 \times 0.325 \\[1em] \text B_\text V = 0.13\ \text G\end{aligned}$$

Magnitude of $\vec{\text B}_\text E$ :

$$\begin{aligned}\text B_\text E = \sqrt{\text B_\text H^2 + \text B_\text V^2} \\[1em] = \sqrt{(0.40)^2 + (0.13)^2} \\[1em] = \sqrt{0.16 + 0.0169} \\[1em] = \sqrt{0.1769} \\[1em] \text B_\text E = 0.42\ \text G\end{aligned}$$

Direction of $\vec{\text B}_\text E$ : $\vec{\text B}_\text E$ lies in the magnetic meridian, and since the north tip of the dip needle points down, it is inclined downward at the angle of dip.

Hence, the earth's magnetic field $\vec{\text B}_\text E$ at the place is 0.42 G. It acts in the magnetic meridian, which is 3.5° west of the geographic meridian, and points 18° below the horizontal.

Given,

• Distance of the neutral point from the centre, r = 10 cm = 0.10 m

• B_H = 0.4 G = 0.4 × 10^-4 T

At a neutral point, the magnetic field due to the magnet is exactly neutralised by the horizontal component of earth's magnetic field.

With the north pole pointing north, the field of the magnet on its axis is directed towards the north, along B_H, so no neutral point can lie there. On the equatorial line, the field of the magnet is directed from N to S, that is, towards the south, opposite to B_H. Hence the neutral point lies on the equatorial line, where

$\frac{\mu_0}{4\pi}\frac{\text m}{\text r^3} = \text B_\text H$

Therefore,

$$\begin{aligned}\text m = \frac{\text B_\text H\thinspace\text r^3}{\left(\frac{\mu_0}{4\pi}\right)} \\[1em] = \frac{0.4 \times 10^{-4} \times (0.10)^3}{10^{-7}} \\[1em] = \frac{0.4 \times 10^{-7}}{10^{-7}} \\[1em] \text m = 0.4\ \text{A m}^2\end{aligned}$$

Hence, the magnetic moment of the magnet is 0.4 A m².

Given,

• Magnetic moment, m = 4.8 A m²

• Distance of the neutral points from the centre, r = 30 cm = 0.30 m

With the north pole pointing towards geographical south, the field of the magnet on its axis is directed from S to N of the magnet, that is, towards the south. This is opposite to B_H, which points towards the north. Hence the neutral points lie on the axis of the magnet, where

$\frac{\mu_0}{4\pi}\frac{2\text m}{\text r^3} = \text B_\text H$

Substituting the values,

$$\begin{aligned}\text B_\text H = 10^{-7} \times \frac{2 \times 4.8}{(0.30)^3} \\[1em] = \frac{9.6 \times 10^{-7}}{0.027} \\[1em] = 3.56 \times 10^{-5}\ \text T \\[1em] \text B_\text H = 0.356\ \text G\end{aligned}$$

Hence, the horizontal component of the earth's magnetic field is 3.56 × 10^-5 T, that is, about 0.36 G.

Note: The textbook prints 0.35 G. The computed value 0.356 G rounds to 0.36 G; the printed figure is truncated.

Given,

• Distance of the neutral points from the centre, r = 12.5 cm

• Earth's magnetic field, B_E = 0.38 G

• Angle of dip, θ = 0°

Since the angle of dip is zero,

$$\begin{aligned}\text B_\text H = \text B_\text E\cos 0^\circ \\[1em] \text B_\text H = 0.38\ \text G\end{aligned}$$

The neutral points lie on the equatorial line, where the field of the magnet (directed from N to S) is opposite to B_H. So the north pole of the magnet points towards the north, and at the neutral points

$\frac{\mu_0}{4\pi}\frac{\text m}{\text r^3} = \text B_\text H = 0.38\ \text G \qquad ...(i)$

(a) Total field on the axis at 12.5 cm :

At an axial point at the same distance r, the field of the magnet is

$$\begin{aligned}\text B_1 = \frac{\mu_0}{4\pi}\frac{2\text m}{\text r^3} \\[1em] = 2 \times 0.38 \\[1em] \text B_1 = 0.76\ \text G\end{aligned}$$

This field is directed from S to N of the magnet, that is, towards the north, along B_H. Therefore, the total field is

$$\begin{aligned}\text B = \text B_1 + \text B_\text H \\[1em] = 0.76 + 0.38 \\[1em] \text B = 1.14\ \text G\end{aligned}$$

(b) Neutral points when the magnet is turned by 180° :

On turning the magnet through 180°, its north pole points towards the south. Now the field of the magnet on its axis is directed towards the south, opposite to B_H, while on the equatorial line it is along B_H. The neutral points therefore shift to the axis. If they are at a distance r' from the centre,

$\frac{\mu_0}{4\pi}\frac{2\text m}{(\text r')^3} = \text B_\text H \qquad ...(ii)$

Comparing Eqs. (i) and (ii),

$$\begin{aligned}\frac{2}{(\text r')^3} = \frac{1}{\text r^3} \\[1em] (\text r')^3 = 2\text r^3 \\[1em] \text r' = 2^{\frac{1}{3}} \times 12.5 \\[1em] = 1.2599 \times 12.5 \\[1em] \text r' = 15.7\ \text{cm}\end{aligned}$$

Hence, (a) the total magnetic field on the axis at 12.5 cm is 1.14 G, and (b) the neutral points lie on the axis of the magnet at 15.7 cm from its centre.

Given,

• B_H = $\sqrt{3}$ B_V

If θ is the angle of dip, then $\tan\theta = \frac{\text B_\text V}{\text B_\text H}$.

$$\begin{aligned}\tan\theta = \frac{\text B_\text V}{\sqrt{3}\ \text B_\text V} \\[1em] = \frac{1}{\sqrt{3}} \\[1em] \theta = 30^\circ\end{aligned}$$

The horizontal component is related to the total field B_E by B_H = B_E cos θ.

$$\begin{aligned}\frac{\text B_\text H}{\text B_\text E} = \cos\theta \\[1em] = \cos 30^\circ \\[1em] \frac{\text B_\text H}{\text B_\text E} = \frac{\sqrt{3}}{2}\end{aligned}$$

Hence, the angle of dip is 30° and the ratio of the horizontal component to the total magnetic field is $\sqrt{3}$ : 2.

Let an electron of mass m_e and charge −e revolve with speed v in a circular orbit of radius r about the nucleus. The revolving electron is a minute current-loop and so behaves like a magnetic dipole.

Current due to the electron : The period of revolution is $\text T = \frac{2\pi\text r}{\text v}$, so the current is

$$\begin{aligned}\text I = \frac{\text e}{\text T} \\[1em] \text I = \frac{\text{ev}}{2\pi\text r} \quad ...(i)\end{aligned}$$

Magnetic dipole moment : The area of the orbit is A = πr². Hence

$$\begin{aligned}\text m = \text{IA} \\[1em] = \frac{\text{ev}}{2\pi\text r} \times \pi\text r^2 \\[1em] \text m = \frac{\text{evr}}{2} \quad ...(ii)\end{aligned}$$

Relation with angular momentum : The orbital angular momentum of the electron is

$\text J = \text m_\text e\text{vr} \quad ...(iii)$

Dividing (ii) by (iii),

$\frac{\text m}{\text J} = \frac{\text e}{2\text m_\text e} \quad ...(iv)$

The ratio $\frac{\text e}{2\text m_\text e}$ is the gyromagnetic ratio of the electron. Since the electron is negatively charged, the current is opposite to its motion, so $\vec{\text m} \space$is directed opposite to $\vec{\text J}$:

$\vec{\text m} = -\frac{\text e}{2\text m_\text e}\vec{\text J}$

In the hydrogen atom, $\text J = \frac{\text{nh}}{2\pi}$, so $\text m = \frac{\text{neh}}{4\pi\text m_\text e}$. For n = 1, m = 9.27 × 10^-24 A m² (the Bohr magneton).

Hence, the magnetic dipole moment of the orbiting electron is $\text m = \frac{\text{evr}}{2} = \frac{\text e}{2\text m_\text e}\text J$, directed opposite to its orbital angular momentum.

Given,

• Angle of dip, θ = 60°
• B_H = 0.4 G

Since the needle rotates in the magnetic meridian, the angle its north tip makes with the horizontal is the angle of dip. The horizontal component is B_H = B_E cos θ.

$$\begin{aligned}\text B_\text E = \frac{\text B_\text H}{\cos\theta} \\[1em] = \frac{0.4}{\cos 60^\circ} \\[1em] = \frac{0.4}{0.5} \\[1em] \text B_\text E = 0.8\ \text G\end{aligned}$$

Hence, the magnitude of the earth's magnetic field at the place is 0.8 G (= 0.8 × 10^-4 T).

The SI unit of magnetic moment is ampere-metre² (A m²), which is the same as joule per tesla (J T^-1).

Magnetic moment is a vector. Its direction is along the magnetic axis of the magnet, from the south pole to the north pole.

Consider a magnetic dipole NS of magnetic length 2l and pole strength q_m, so that its magnetic moment is M = q_m × 2l. Let P be a point on its axis at a distance x from the centre O, on the side of the north pole.

Distance of P from N = (x − l), and from S = (x + l).

Field due to the north pole : directed away from N, along NP,

$\text B_\text N = \frac{\mu_0}{4\pi}\frac{\text q_\text m}{(\text x - \text l)^2} \quad ...(i)$

Field due to the south pole : directed towards S, along PS,

$\text B_\text S = \frac{\mu_0}{4\pi}\frac{\text q_\text m}{(\text x + \text l)^2} \quad ...(ii)$

Since B_N > B_S, the resultant field at P is along NP, i.e., in the direction S → N:

$$\begin{aligned}\text B = \text B_\text N - \text B_\text S \\[1em] = \frac{\mu_0}{4\pi}\text q_\text m\left[\frac{1}{(\text x - \text l)^2} - \frac{1}{(\text x + \text l)^2}\right] \\[1em] = \frac{\mu_0}{4\pi}\text q_\text m\left[\frac{(\text x + \text l)^2 - (\text x - \text l)^2}{(\text x^2 - \text l^2)^2}\right] \\[1em] = \frac{\mu_0}{4\pi}\frac{\text q_\text m \times 4\text{xl}}{(\text x^2 - \text l^2)^2} \\[1em] = \frac{\mu_0}{4\pi}\frac{2(\text q_\text m \times 2\text l)\text x}{(\text x^2 - \text l^2)^2} \\[1em] \text B = \frac{\mu_0}{4\pi}\frac{2\text{Mx}}{(\text x^2 - \text l^2)^2} \quad ...(iii)\end{aligned}$$

For a short dipole (l << x), l² may be neglected in comparison with x²:

$\text B = \frac{\mu_0}{4\pi}\frac{2\text M}{\text x^3}$

Hence, the magnetic field at an axial point is $\text B = \frac{\mu_0}{4\pi}\frac{2\text{Mx}}{(\text x^2 - \text l^2)^2}$, directed along the axis from the south pole to the north pole.

Given,

• Charge on helium nucleus, q = 2e = 3.2 × 10^-19 C
• Radius of the circle, r = 0.8 m
• Time for one revolution, T = 2 s

The revolving nucleus is equivalent to a circular current-loop carrying current $\text I = \frac{\text q}{\text T}$.

$$\begin{aligned}\text I = \frac{3.2 \times 10^{-19}}{2} \\[1em] \text I = 1.6 \times 10^{-19}\ \text A\end{aligned}$$

The magnetic field at the centre of a circular loop is $\text B = \frac{\mu_0\text I}{2\text r}$.

$$\begin{aligned}\text B = \frac{4\pi \times 10^{-7} \times 1.6 \times 10^{-19}}{2 \times 0.8} \\[1em] = 4\pi \times 10^{-26} \\[1em] \text B = 1.256 \times 10^{-25}\ \text T\end{aligned}$$

Hence, the magnetic field at the centre of the circle is 1.256 × 10^-25 T.

45°.

Explanation:
The angle of dip θ is given by $\tan\theta = \frac{\text B_\text V}{\text B_\text H}$. When B_V = B_H, tan θ = 1, so θ = 45°.

$\sqrt{3}$ B.

Explanation:
From $\tan\theta = \frac{\text B_\text V}{\text B_\text H}$, with B_H = B and θ = 60°,

$$\begin{aligned}\text B_\text V = \text B_\text H\tan\theta \\[1em] = \text B\tan 60^\circ \\[1em] \text B_\text V = \sqrt{3}\ \text B\end{aligned}$$

Ampere-metre² (A m²), or equivalently joule per tesla (J T^-1).

45°

Reason — The angle of dip θ satisfies $\tan\theta = \frac{\text B_\text V}{\text B_\text H}$. With B_V = B_H, tan θ = 1, so θ = 45°.

The angle of dip at a place is the angle which the direction of the earth's magnetic field makes with the horizontal, in the magnetic meridian at that place.

None of the given options is exactly correct; W ≈ 9.48 μJ

Reason — Work done in turning the coil through 180° from its stable position is W = mB(cos 0° − cos 180°) = 2mB, where m = NIA.

$$\begin{aligned}\text W = 2\text{NIAB} \\[1em] = 2 \times 250 \times 85 \times 10^{-6} \times \left(2.1 \times 10^{-2} \times 1.25 \times 10^{-2}\right) \times 0.85 \\[1em] \approx 9.48 \times 10^{-6}\ \text J \\[1em] \text W \approx 9.48\ \mu\text J\end{aligned}$$

Note: The textbook keys 9.1 μJ because of an arithmetic slip in its printed working; 9.1 μJ is only the option nearest to the correct value, 9.48 μJ.

6.65 s

Reason — The needle performs S.H.M. with period $\text T = 2\pi\sqrt{\frac{\text I}{\text{mB}}}$, where I is its moment of inertia.

$$\begin{aligned}\text T = 2\pi\sqrt{\frac{7.5 \times 10^{-6}}{6.7 \times 10^{-2} \times 0.01}} \\[1em] = 2\pi \times 0.1058 \\[1em] \text T = 0.665\ \text s\end{aligned}$$

Time for 10 oscillations = 10T = 6.65 s.

cot² θ = cot² θ₁ + cot² θ₂

Reason — If one plane makes angle α with the magnetic meridian, the other makes (90° − α). Using $\tan\theta' = \frac{\tan\theta}{\cos\alpha}$ for each plane,

$$\begin{aligned}\tan\theta_1 = \frac{\tan\theta}{\cos\alpha} \\[1em] \Rightarrow \cos\alpha = \frac{\cot\theta_1}{\cot\theta} \\[1em] \tan\theta_2 = \frac{\tan\theta}{\sin\alpha} \\[1em] \Rightarrow \sin\alpha = \frac{\cot\theta_2}{\cot\theta} \\[1em] \cos^2\alpha + \sin^2\alpha = 1 \\[1em] \Rightarrow \cot^2\theta_1 + \cot^2\theta_2 = \cot^2\theta\end{aligned}$$

$\sqrt{2}$

Reason — m = IπR². Doubling m at constant I makes the new radius R′ = $\sqrt{2}$ R. Since $\text B = \frac{\mu_0\text I}{2\text R} \propto \frac{1}{\text R}$,

$$\begin{aligned}\frac{\text B_1}{\text B_2} = \frac{\text R'}{\text R} \\[1em] \frac{\text B_1}{\text B_2} = \sqrt{2}\end{aligned}$$

$\frac{4\text m}{\pi}$

Reason — For the square of side a, m = Ia² and the wire length is 4a. Bent into a circle, 2πr = 4a, so $\text r = \frac{2\text a}{\pi}$.

$$\begin{aligned}\text m' = \text I\pi\text r^2 \\[1em] = \text I\pi\left(\frac{2\text a}{\pi}\right)^2 \\[1em] = \frac{4\text{Ia}^2}{\pi} \\[1em] \text m' = \frac{4\text m}{\pi}\end{aligned}$$

0

Reason — P is an axial point of X, so the field of X at P is $\left(\frac{\mu_0}{4\pi}\right)\frac{2\text m}{\left(\frac{\text d}{2}\right)^3}$, horizontal, from S to N of X. P is an equatorial point of Y (moment 2m), so the field of Y at P is $\left(\frac{\mu_0}{4\pi}\right)\frac{2\text m}{\left(\frac{\text d}{2}\right)^3}$, opposite to the moment of Y, i.e., vertically upward. The two equal, perpendicular fields give a resultant at 45° to the horizontal, which is along the velocity of the particle. Hence F = qvB sin 0° = 0.

0.27 Nm

Reason — m = NIA = 100 × 3 × (5 × 10^-2 × 2.5 × 10^-2) = 0.375 A m². In the X-Z plane, $\vec{\text m} \space$is along the Y-axis; after tilting through 45° about the Z-axis it makes 45° with $\vec{\text B} \space$(X-axis).

$$\begin{aligned}\tau = \text{mB}\sin 45^\circ \\[1em] = 0.375 \times 1 \times 0.707 \\[1em] \tau \approx 0.27\ \text{Nm}\end{aligned}$$

6.5 × 10^-5 N

Reason — The freely suspended needle rests along the earth's field, so the angle of dip is 45° and B_V = B_H tan 45° = 18 × 10^-6 T. When the needle is horizontal, B_H acts along it (no torque), while B_V acts on the poles (strength q_m) to give a couple q_mB_V × l. Balancing it with a vertical force F at one end, about the mid-point,

$$\begin{aligned}\text F \times \frac{\text l}{2} = \text q_\text m\text B_\text V\text l \\[1em] \text F = 2\text q_\text m\text B_\text V \\[1em] = 2 \times 1.8 \times 18 \times 10^{-6} \\[1em] \text F \approx 6.5 \times 10^{-5}\ \text N\end{aligned}$$

$\frac{\pi}{4}\text n\rho\text l^3$

Reason — An element dx at distance x carries charge $\text{dq} = \rho_0\frac{\text x}{\text l}\thinspace\text{dx}$. Rotating n times per second, it forms a loop of current n dq and area πx², so $\text{dm} = \frac{\text n\pi\rho_0}{\text l}\text x^3\thinspace\text{dx}$.

$$\begin{aligned}\text m = \int_0^\text l \frac{\text n\pi\rho_0}{\text l}\text x^3\thinspace\text{dx} \\[1em] = \frac{\text n\pi\rho_0}{\text l}\left[\frac{\text x^4}{4}\right]_0^\text l \\[1em] = \frac{\text n\pi\rho_0}{\text l} \times \frac{\text l^4}{4} \\[1em] \text m = \frac{\pi}{4}\text n\rho_0\text l^3\end{aligned}$$

Here ρ in the options stands for ρ₀.

0.7

Reason — A compass needle oscillates in the horizontal plane under B_H = B cos θ, so $\text T = 2\pi\sqrt{\frac{\text I}{\text{mB}\cos\theta}}$ and the frequency f satisfies f² ∝ B cos θ.

$$\begin{aligned}\frac{\text f_1^2}{\text f_2^2} = \frac{\text B_1\cos 45^\circ}{\text B_2\cos 30^\circ} \\[1em] \frac{\text B_1}{\text B_2} = \left(\frac{30}{40}\right)^2 \times \frac{\cos 30^\circ}{\cos 45^\circ} \\[1em] = \frac{9}{16} \times \sqrt{\frac{3}{2}} \\[1em] \frac{\text B_1}{\text B_2} \approx 0.7\end{aligned}$$

A is located in the northern hemisphere and B is located in the southern hemisphere.

Reason — The angle of dip is zero at the magnetic equator, positive in the northern hemisphere (north end of the needle dips down) and negative in the southern hemisphere (south end dips down).

7.2 × 10^-2 J

Reason — From τ = mB sin θ, $\text m = \frac{0.018}{0.06 \times \sin 30^\circ}$ = 0.6 A m². Work from stable (θ = 0°) to unstable (θ = 180°) equilibrium is W = mB(cos 0° − cos 180°) = 2mB.

$$\begin{aligned}\text W = 2 \times 0.6 \times 0.06 \\[1em] \text W = 7.2 \times 10^{-2}\ \text J\end{aligned}$$

$\frac{\text{IL}^2}{4\pi}$ Am²

Reason — The circumference is L = 2πR, so $\text R = \frac{\text L}{2\pi}$.

$$\begin{aligned}\text M = \text I\pi\text R^2 \\[1em] = \text I\pi\left(\frac{\text L}{2\pi}\right)^2 \\[1em] \text M = \frac{\text{IL}^2}{4\pi}\end{aligned}$$

$\sqrt{2}\text{abI}$, along $\left(\frac{\hat{\text j}}{\sqrt{2}} + \frac{\hat{\text k}}{\sqrt{2}}\right)$

Reason — The loop ABCDEFA is equivalent to loops ABCDA and ADEFA (the currents along AD and DA cancel), each of area ab carrying current I. By the right-hand rule, ABCDA (in the X-Y plane) has moment abI $\hat{\text k}$ and ADEFA (in the X-Z plane) has moment abI $\hat{\text j}$.

$$\begin{aligned}\vec{\text M} = \text{abI}\thinspace\hat{\text j} + \text{abI}\thinspace\hat{\text k} \\[1em] \vec{\text M} = \sqrt{2}\text{abI}\left(\frac{\hat{\text j}}{\sqrt{2}} + \frac{\hat{\text k}}{\sqrt{2}}\right)\end{aligned}$$

$-\frac{\text{mv}^2\vec{\text B}}{2\text B^2}$

Reason — The particle moves in a circle of radius $\text R = \frac{\text{mv}}{\text{qB}}$, forming a current $\frac{\text{qv}}{2\pi\text R}$. Its magnetic moment is $\frac{\text{qv}}{2\pi\text R} \times \pi\text R^2 = \frac{\text{qvR}}{2}$, directed opposite to $\vec{\text B}$.

$$\begin{aligned}\vec{\text M} = \frac{\text{qv}}{2}\left(\frac{\text{mv}}{\text{qB}}\right)\left(-\frac{\vec{\text B}}{\text B}\right) \\[1em] \vec{\text M} = -\frac{\text{mv}^2\vec{\text B}}{2\text B^2}\end{aligned}$$

$\sqrt{3}\ \text{Ia}^2$ and 3 Ia²

Reason — Triangle: each turn uses 3a, so N = $\frac{12\text a}{3\text a}$ = 4 and area = $\frac{\sqrt{3}}{4}\text a^2$, giving m₁ = NIA = $4\text I \times \frac{\sqrt{3}}{4}\text a^2 = \sqrt{3}\ \text{Ia}^2$. Square: each turn uses 4a, so N = 3 and m₂ = 3 × I × a² = 3 Ia².

2.880 JT^-1

Reason — With the north pole pointing geographic north, the neutral points lie on the equatorial line, where the field of the magnet equals B_H. Here l = 7 cm = 0.07 m, r = 0.18 m and B_H = 0.4 × 10^-4 T.

$$\begin{aligned}\frac{\mu_0}{4\pi}\frac{\text M}{(\text r^2 + \text l^2)^{\frac{3}{2}}} = \text B_\text H \\[1em] \text M = \frac{0.4 \times 10^{-4} \times \left(0.18^2 + 0.07^2\right)^{\frac{3}{2}}}{10^{-7}} \\[1em] = \frac{0.4 \times 10^{-4} \times 7.204 \times 10^{-3}}{10^{-7}} \\[1em] \text M \approx 2.880\ \text J\ \text T^{-1}\end{aligned}$$

$\tan^{-1}\frac{\sqrt{3}}{2}$

Reason — From $\tan\theta' = \frac{\tan\theta}{\cos\alpha}$ with θ′ = 45° and α = 30°,

$$\begin{aligned}\tan\theta = \tan 45^\circ \times \cos 30^\circ \\[1em] \tan\theta = \frac{\sqrt{3}}{2}\end{aligned}$$

True dip is less than the apparent dip.

Reason — Apparent dip θ′ and true dip θ are related by $\tan\theta' = \frac{\tan\theta}{\cos\alpha}$. Since cos α < 1 in any plane other than the meridian, tan θ′ > tan θ, i.e., θ′ > θ.

1 × 10^-4 T

Reason — From $\tan\theta = \frac{\text B_\text V}{\text B_\text H}$, $\text B_\text H = \frac{6 \times 10^{-5}}{\frac{3}{4}}$ = 8 × 10^-5 T.

$$\begin{aligned}\text B_\text E = \sqrt{\text B_\text H^2 + \text B_\text V^2} \\[1em] = \sqrt{(8 \times 10^{-5})^2 + (6 \times 10^{-5})^2} \\[1em] \text B_\text E = 1 \times 10^{-4}\ \text T\end{aligned}$$

8 : 3

Reason — From $\text T = 2\pi\sqrt{\frac{\text I}{\text{MB}}}$, M ∝ $\frac{\text I}{\text T^2}$ in the same field.

$$\begin{aligned}\frac{\text M_1}{\text M_2} = \frac{\text I_1}{\text I_2} \times \left(\frac{\text T_2}{\text T_1}\right)^2 \\[1em] = \frac{3}{2} \times \left(\frac{4}{3}\right)^2 \\[1em] \frac{\text M_1}{\text M_2} = \frac{8}{3}\end{aligned}$$

$\tan^{-1}\left(\sqrt{\frac{3}{2}}\right)$

Reason — From $\tan\theta' = \frac{\tan\theta}{\cos\alpha}$ with θ′ = 60° and α = 45°,

$$\begin{aligned}\tan\theta = \tan 60^\circ \times \cos 45^\circ \\[1em] = \sqrt{3} \times \frac{1}{\sqrt{2}} \\[1em] \tan\theta = \sqrt{\frac{3}{2}}\end{aligned}$$

$\sqrt{3}$ : 4

Reason — The needle oscillates under B_H = B cos θ, so f² ∝ B cos θ.

$$\begin{aligned}\frac{\text B_\text Q\cos 60^\circ}{\text B_\text P\cos 30^\circ} = \left(\frac{10}{20}\right)^2 \\[1em] \frac{\text B_\text Q}{\text B_\text P} = \frac{1}{4} \times \frac{\cos 30^\circ}{\cos 60^\circ} \\[1em] = \frac{1}{4} \times \sqrt{3} \\[1em] \frac{\text B_\text Q}{\text B_\text P} = \frac{\sqrt{3}}{4}\end{aligned}$$

$\frac{1}{\sqrt{3}}$

Reason — From B_H = B_E cos θ,

$$\begin{aligned}\text B_\text E = \frac{0.5}{\cos 30^\circ} \\[1em] = \frac{0.5}{\frac{\sqrt{3}}{2}} \\[1em] \text B_\text E = \frac{1}{\sqrt{3}}\ \text G\end{aligned}$$

1280π²

Reason — Time period $\text T = \frac{5}{20}$ = 0.25 s. From $\text T = 2\pi\sqrt{\frac{\text I}{\text{mB}}}$, $\text m = \frac{4\pi^2\text I}{\text B\text T^2}$.

$$\begin{aligned}\text m = \frac{4\pi^2 \times 9.8 \times 10^{-6}}{0.049 \times (0.25)^2} \\[1em] = 4\pi^2 \times 3.2 \times 10^{-3} \\[1em] \text m = 1280\pi^2 \times 10^{-5}\ \text A\ \text m^2\end{aligned}$$

Comparing with x × 10^-5 A m², x = 1280π².

Given,

• m₁ = m₂ = 1 A m²
• OP = r = 1 m

From the figure, m₁ at O points perpendicular to OP, and m₂ at P points along OP. So P lies on the equatorial (broad-side-on) line of m₁, where the field of m₁ is $\text B_1 = \frac{\mu_0}{4\pi}\frac{\text m_1}{\text r^3}$, directed opposite to $\vec{\text m}_1$.

$$\begin{aligned}\text B_1 = 10^{-7} \times \frac{1}{1^3} \\[1em] \text B_1 = 1 \times 10^{-7}\ \text T\end{aligned}$$

$\vec{\text B}_1$ is perpendicular to $\vec{\text m}_2$, so the torque on m₂ is τ = m₂B₁ sin 90°.

$$\begin{aligned}\tau = 1 \times 1 \times 10^{-7} \times 1 \\[1em] \tau = 1 \times 10^{-7}\ \text{Nm}\end{aligned}$$

Hence, the torque is 1 × 10^-7 Nm, so the required value is 1.

Given,

• Magnetic moment, M = 9.85 × 10^-2 A m²
• Moment of inertia, I = 5 × 10^-6 kg m²
• Time period, $\text T = \frac{5}{10}$ = 0.5 s
• π² = 9.85

For a magnetic needle oscillating in a uniform field, $\text T = 2\pi\sqrt{\frac{\text I}{\text{MB}}}$, so $\text B = \frac{4\pi^2\text I}{\text M\text T^2}$.

$$\begin{aligned}\text B = \frac{4 \times 9.85 \times 5 \times 10^{-6}}{9.85 \times 10^{-2} \times (0.5)^2} \\[1em] = \frac{2 \times 10^{-5}}{2.5 \times 10^{-3}} \\[1em] = 8 \times 10^{-3}\ \text T \\[1em] \text B = 8\ \text{mT}\end{aligned}$$

Hence, the magnitude of the magnetic field is 8 mT.

Note: The unit of magnetic moment printed in the question as A/m² should read A m².

Given,

• f₁ = 20 oscillations per minute, θ₁ = 30°
• f₂ = 30 oscillations per minute, θ₂ = 60°

The compass needle oscillates in the horizontal plane under B_H = B cos θ, so $\text T = 2\pi\sqrt{\frac{\text I}{\text{mB}\cos\theta}}$ and f² ∝ B cos θ.

$$\begin{aligned}\frac{\text f_1^2}{\text f_2^2} = \frac{\text B_1\cos 30^\circ}{\text B_2\cos 60^\circ} \\[1em] \frac{\text B_1}{\text B_2} = \left(\frac{20}{30}\right)^2 \times \frac{\cos 60^\circ}{\cos 30^\circ} \\[1em] = \frac{4}{9} \times \frac{1}{\sqrt{3}} \\[1em] = \frac{4}{9\sqrt{3}} \\[1em] \frac{\text B_1}{\text B_2} = \frac{4}{\sqrt{243}}\end{aligned}$$

Comparing with $\frac{4}{\sqrt{\text x}}$,

Hence, x = 243.

Given,

• Magnetic potential, V = 1.5 × 10^-5 T m
• Distance from centre, r = 20 cm = 0.2 m
• $\frac{\mu_0}{4\pi}$ = 10^-7 T m A^-1

The magnetic potential at an axial point of a short magnetic dipole is $\text V = \frac{\mu_0}{4\pi}\frac{\text M}{\text r^2}$, so $\text M = \frac{\text V\text r^2}{\left(\frac{\mu_0}{4\pi}\right)}$.

$$\begin{aligned}\text M = \frac{1.5 \times 10^{-5} \times (0.2)^2}{10^{-7}} \\[1em] = \frac{6 \times 10^{-7}}{10^{-7}} \\[1em] \text M = 6\ \text A\ \text m^2\end{aligned}$$

Hence, the magnetic moment of the dipole is 6 A m².