Question 1

1 A short bar magnet placed with its axis at $30^{\circ}$ with a uniform external magnetic field of 0.25 T experiences a torque of magnitude equal to $4.5 	imes 10^{-2} \mathrm{~J}$. What is the magnitude of magnetic moment of the magnet?

Accepted Answer

Given: Torque $ \tau = 4.5 \times 10^{-2} \, \mathrm{J} $, Magnetic field $ B = 0.25 \, \mathrm{T} $, Angle $ \theta = 30^{\circ} $.

Torque on a magnetic dipole in a magnetic field is given by:
$$
\tau = m B \sin \theta
$$
where $ m $ is the magnetic moment.

Rearranging for $ m $:
$$
m = \frac{\tau}{B \sin \theta}
$$

Calculate $ \sin 30^{\circ} = 0.5 $.

Therefore,
$$
m = \frac{4.5 \times 10^{-2}}{0.25 \times 0.5} = \frac{4.5 \times 10^{-2}}{0.125} = 0.36 \, \mathrm{A \cdot m^2}
$$

Hence, the magnitude of the magnetic moment is $0.36 \, \mathrm{A \cdot m^2}$. — The torque on a magnetic dipole in a magnetic field is $ \tau = m B \sin \theta $. Given $ \tau $, $ B $, and $ \theta $, we solve for $ m $ by dividing the torque by the product $ B \sin \theta $. Calculating $ \sin 30^{\circ} = 0.5 $, we substitute and find $ m = 0.36 \, \mathrm{A \cdot m^2} $.

Question 2

2 A short bar magnet of magnetic moment $m = 0.32 \, \mathrm{JT}^{-1}$ is placed in a uniform magnetic field of 0.15 T. If the bar is free to rotate in the plane of the field, which orientation would correspond to its (a) stable, and (b) unstable equilibrium? What is the potential energy of the magnet in each case?

Accepted Answer

(a) Stable equilibrium occurs when the magnetic moment $ \vec{m} $ is aligned with the magnetic field $ \vec{B} $, i.e., the angle $ \theta = 0^{\circ} $.

(b) Unstable equilibrium occurs when $ \vec{m} $ is opposite to $ \vec{B} $, i.e., $ \theta = 180^{\circ} $.

Potential energy of a magnetic dipole in a magnetic field is given by:
$$
U = -m B \cos \theta
$$

Calculate potential energy for each case:

For stable equilibrium ($ \theta = 0^{\circ} $):
$$
U = -m B \cos 0^{\circ} = -m B = -0.32 \times 0.15 = -0.048 \, \mathrm{J}
$$

For unstable equilibrium ($ \theta = 180^{\circ} $):
$$
U = -m B \cos 180^{\circ} = -m B (-1) = +m B = +0.048 \, \mathrm{J}
$$

Hence, the stable equilibrium corresponds to minimum potential energy $-0.048 \, \mathrm{J}$, and unstable equilibrium corresponds to maximum potential energy $+0.048 \, \mathrm{J}$. — The magnetic dipole aligns itself to minimize potential energy. When aligned with the field ($ \theta=0^{\circ} $), potential energy is minimum (stable). When opposite ($ \theta=180^{\circ} $), potential energy is maximum (unstable). Using $ U = -m B \cos \theta $, substituting values gives the energies for both cases.

Question 3

3 A closely wound solenoid of 800 turns and area of cross section $2.5 	imes 10^{-4} \, \mathrm{m}^2$ carries a current of 3.0 A. Explain the sense in which the solenoid acts like a bar magnet. What is its associated magnetic moment?

Accepted Answer

A solenoid carrying current produces a magnetic field similar to that of a bar magnet, with a north and south pole at its ends. The magnetic field lines inside the solenoid are nearly uniform and parallel, resembling the field of a bar magnet.

Magnetic moment $ m $ of a current-carrying coil (or solenoid) is given by:
$$
m = N I A
$$
where
$ N = 800 $ (number of turns),
$ I = 3.0 \, \mathrm{A} $ (current),
$ A = 2.5 \times 10^{-4} \, \mathrm{m}^2 $ (area of cross-section).

Calculate:
$$
m = 800 \times 3.0 \times 2.5 \times 10^{-4} = 800 \times 7.5 \times 10^{-4} = 0.6 \, \mathrm{A \cdot m^2}
$$

Hence, the magnetic moment of the solenoid is $0.6 \, \mathrm{A \cdot m^2}$. — The solenoid acts like a bar magnet because the current loops produce a magnetic field with distinct poles. The magnetic moment is the product of number of turns, current, and area of each turn. Substituting the given values yields the magnetic moment.

Question 4

5 A bar magnet of magnetic moment 1.5 J T⁻¹ lies aligned with the direction of a uniform magnetic field of 0.22 T.

(a) What is the amount of work required by an external torque to turn the magnet so as to align its magnetic moment: (i) normal to the field direction, (ii) opposite to the field direction?

(b) What is the torque on the magnet in cases (i) and (ii)?

Accepted Answer

Given:
Magnetic moment, m = 1.5 J T⁻¹
Magnetic field, B = 0.22 T

(a) Work done by external torque to rotate the magnet from initial alignment to a new angle θ with the field is given by:
W = mB (cos θ_initial - cos θ_final)

(i) To align magnetic moment normal (90°) to the field:
Initial angle θ_initial = 0° (aligned)
Final angle θ_final = 90°

W = mB (cos 0° - cos 90°) = 1.5 × 0.22 × (1 - 0) = 0.33 J

(ii) To align magnetic moment opposite (180°) to the field:
θ_final = 180°

W = mB (cos 0° - cos 180°) = 1.5 × 0.22 × (1 - (-1)) = 1.5 × 0.22 × 2 = 0.66 J

(b) Torque τ on a magnetic dipole in a magnetic field is:
τ = mB sin θ

(i) When magnetic moment is normal to field (θ = 90°):
τ = 1.5 × 0.22 × sin 90° = 0.33 N·m

(ii) When magnetic moment is opposite to field (θ = 180°):
sin 180° = 0

τ = 0 N·m — Step 1: Understand that work done by external torque changes potential energy of the magnetic dipole.
Step 2: Use formula for work done W = mB (cos θ_initial - cos θ_final).
Step 3: Calculate for θ_final = 90° and 180°.
Step 4: Calculate torque τ = mB sin θ for both cases.
Step 5: Note that torque is zero when magnetic moment is aligned or anti-aligned with the field (θ = 0° or 180°).

Question 5

6 A closely wound solenoid of 2000 turns and area of cross-section 1.6 × 10⁻⁴ m², carrying a current of 4.0 A, is suspended through its centre allowing it to turn in a horizontal plane.

(a) What is the magnetic moment associated with the solenoid?

(b) What is the force and torque on the solenoid if a uniform horizontal magnetic field of 7.5 × 10⁻² T is set up at an angle of 30° with the axis of the solenoid?

Accepted Answer

Given:
Number of turns, N = 2000
Area of cross-section, A = 1.6 × 10⁻⁴ m²
Current, I = 4.0 A
Magnetic field, B = 7.5 × 10⁻² T
Angle between field and solenoid axis, θ = 30°

(a) Magnetic moment, m = NIA
m = 2000 × 4.0 × 1.6 × 10⁻⁴ = 1.28 A·m²

(b) Force on a magnetic dipole in uniform magnetic field is zero:
F = 0

Torque, τ = mB sin θ
= 1.28 × 7.5 × 10⁻² × sin 30°
= 1.28 × 0.075 × 0.5 = 0.048 N·m — Step 1: Calculate magnetic moment using m = NIA.
Step 2: Recognize that force on magnetic dipole in uniform field is zero.
Step 3: Calculate torque using τ = mB sin θ.
Step 4: Substitute values and compute numerical results.

Question 6

7 A short bar magnet has a magnetic moment of 0.48 J T⁻¹. Give the direction and magnitude of the magnetic field produced by the magnet at a distance of 10 cm from the centre of the magnet on (a) the axis, (b) the equatorial lines (normal bisector) of the magnet.

Accepted Answer

Given:
Magnetic moment, m = 0.48 J T⁻¹
Distance, r = 10 cm = 0.10 m

Magnetic field on the axis of a dipole:
B_axis = \frac{\mu_0}{4\pi} \frac{2m}{r^3}

Magnetic field on the equatorial line:
B_equatorial = \frac{\mu_0}{4\pi} \frac{m}{r^3}

Where \mu_0 = 4\pi 	imes 10^{-7} T·m/A

Calculate r³:
r³ = (0.10)^3 = 1.0 × 10^{-3} m³

Calculate B_axis:
B_axis = (10^{-7}) 	imes \frac{2 	imes 0.48}{1.0 	imes 10^{-3}} = 10^{-7} 	imes 960 = 9.6 	imes 10^{-5} T

Calculate B_equatorial:
B_equatorial = (10^{-7}) 	imes \frac{0.48}{1.0 	imes 10^{-3}} = 4.8 	imes 10^{-5} T

Directions:
(a) On the axis, magnetic field is along the axis of the magnet, away from the north pole.
(b) On the equatorial line, magnetic field is perpendicular to the axis of the magnet and directed opposite to the magnetic moment. — Step 1: Use formulas for magnetic field due to dipole on axis and equatorial line.
Step 2: Substitute values for m and r.
Step 3: Calculate numerical values.
Step 4: State directions based on dipole field lines.

Question 7

Which of the following statements about the Earth’s magnetic field is true?

Accepted Answer

The magnetic field points approximately from geographic south to north — The Earth behaves like a giant magnet with its magnetic field directed approximately from the geographic south pole to the geographic north pole, which is why compass needles point north.

Question 8

Identify the structure labelled as the 'North Pole' in Figure 5.1, which shows iron filings arranged around a bar magnet.

Accepted Answer

The tip of the bar magnet that points approximately towards the geographic north when freely suspended — The north pole of a bar magnet is defined as the pole that points towards the geographic north when the magnet is freely suspended. In Figure 5.1, this corresponds to the tip where iron filings converge and the magnetic field lines emerge.

Question 9

Which property of magnetic field lines is NOT true?

Accepted Answer

They begin at the north pole and end at the south pole — Unlike electric field lines, magnetic field lines do not begin or end on poles; they form continuous closed loops. This is because isolated magnetic monopoles do not exist.

Question 10

Explain why breaking a bar magnet into two halves results in two smaller magnets each having a north and south pole.

Accepted Answer

When a bar magnet is broken into two pieces, each piece behaves like a smaller magnet with both north and south poles. This is because magnetic poles always come in pairs, and isolated magnetic monopoles do not exist. The magnetic domains in each piece align to form complete dipoles. — Magnetic monopoles have never been observed; thus, breaking a magnet does not isolate poles but creates smaller dipoles. Each half contains aligned magnetic domains forming north and south poles.

Question 11

The magnitude of the magnetic field $B$ at a point $P$ on the axis of a finite solenoid is given by $B = \frac{\mu_0}{4\pi} \frac{2m}{r^3}$. What does $m$ represent in this expression?

Accepted Answer

Magnetic moment of the solenoid — In the formula, $m$ represents the magnetic moment of the solenoid, which is analogous to the magnetic moment of a bar magnet producing the same magnetic field at large distances.

Question 12

Describe the torque experienced by a magnetic dipole of moment $\mathbf{m}$ placed in a uniform magnetic field $\mathbf{B}$ at an angle $	heta$ between $\mathbf{m}$ and $\mathbf{B}$.

Accepted Answer

The magnetic dipole experiences a torque $	au = mB \sin 	heta$ that tends to align the dipole moment with the magnetic field. The torque is maximum when $	heta = 90^\circ$ and zero when $	heta = 0^\circ$ or $180^\circ$. For example, a compass needle aligns along Earth's magnetic field due to this torque. — The torque arises from the interaction of the magnetic moment with the external field, causing rotational motion to minimize potential energy. The formula $	au = mB \sin 	heta$ quantifies this effect.

Question 13

What is the expression for the magnetic potential energy $U_m$ of a magnetic dipole with moment $\mathbf{m}$ in a magnetic field $\mathbf{B}$, and what is its physical significance?

Accepted Answer

The magnetic potential energy is $U_m = - \mathbf{m} \cdot \mathbf{B} = - m B \cos 	heta$. It represents the energy stored due to the orientation of the dipole in the magnetic field. The energy is minimum when the dipole is aligned with the field ($	heta = 0^\circ$) and maximum when anti-aligned ($	heta = 180^\circ$). — This energy expression shows that the dipole tends to align with the magnetic field to minimize potential energy, explaining stable and unstable equilibrium positions.

Question 14

In the electrostatic analogy for magnetic dipoles, which of the following substitutions is correct?

Accepted Answer

$\mathbf{E} \rightarrow \mathbf{B}$, $\mathbf{p} \rightarrow \mathbf{m}$, $\frac{1}{4 \pi \varepsilon_0} \rightarrow \frac{\mu_0}{4 \pi}$ — The analogy between electric and magnetic dipoles involves replacing the electric field $\mathbf{E}$ by magnetic field $\mathbf{B}$, the electric dipole moment $\mathbf{p}$ by the magnetic dipole moment $\mathbf{m}$, and the electrostatic constant $1/(4 \pi \varepsilon_0)$ by the magnetic constant $\mu_0/(4 \pi)$.

Question 15

Identify the configuration of two magnetised needles shown in Figure 5.4 that corresponds to unstable equilibrium.

Accepted Answer

The configuration where the magnetic moment of needle Q is anti-parallel to the magnetic field due to needle P, such as $\mathbf{Q}_2$ in the figure. — Unstable equilibrium occurs when the magnetic moment is opposite to the magnetic field direction, causing the system to move away from this state upon slight perturbation. Figure 5.4 shows $\mathbf{Q}_2$ with the moment anti-parallel to $\mathbf{B}_P$.

Magnetism and Matter

Magnetism and Matter — Study Notes

5.1 INTRODUCTION

5.2 THE BAR MAGNET

5.2.4 The electrostatic analog

Practice Questions — Magnetism and Matter

All 8 Chapters in Physics Part-I