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Magnetism and Matter
Physics XII - Chapter 05: Complete NCERT Exercise Solutions
Master Magnetism and Matter with NCERT solutions for magnetic materials, Earth's magnetism, magnetic properties, and hysteresis.
Question 5.1
Answer the following questions regarding earth's magnetism:
(a) A vector needs three quantities for its specification. Name the three independent quantities conventionally used to specify the earth's magnetic field.
(b) The angle of dip at a location in southern India is about \(18^\circ\). Would you expect a greater or smaller dip angle in Britain?
(c) If you made a map of magnetic field lines at Melbourne in Australia, would the lines seem to go into the ground or come out of the ground?
(d) In which direction would a compass free to move in the vertical plane point to, if located right on the geomagnetic north or south pole?
(e) The earth's field, it is claimed, roughly approximates the field due to a dipole of magnetic moment \(8 \times 10^{22} \, \text{J T}^{-1}\) located at its centre. Check the order of magnitude of this number in some way.
(f) Geologists claim that besides the main magnetic N-S poles, there are several local poles on the earth's surface oriented in different directions. How is such a thing possible at all?
Answer & Explanation:
(a) Three quantities: Magnetic declination, angle of dip (inclination), horizontal component of earth's magnetic field.
(b) Britain: Greater dip angle (closer to magnetic pole).
(c) Melbourne: Field lines come out of the ground (southern hemisphere).
(d) At geomagnetic pole: Compass points vertically downward (north pole) or upward (south pole).
(e) Order check: Compare with bar magnet moment (~1 J/T). Earth’s moment is enormous (~10²³ times larger), plausible for planetary scale.
(f) Local poles: Due to magnetic anomalies from mineral deposits, rock magnetism, etc.
(b) Britain: Greater dip angle (closer to magnetic pole).
(c) Melbourne: Field lines come out of the ground (southern hemisphere).
(d) At geomagnetic pole: Compass points vertically downward (north pole) or upward (south pole).
(e) Order check: Compare with bar magnet moment (~1 J/T). Earth’s moment is enormous (~10²³ times larger), plausible for planetary scale.
(f) Local poles: Due to magnetic anomalies from mineral deposits, rock magnetism, etc.
Question 5.2
Answer the following questions:
(a) The earth's magnetic field varies from point to point in space. Does it also change with time? If so, on what time scale does it change appreciably?
(b) The earth's core is known to contain iron. Yet geologists do not regard this as a source of the earth's magnetism. Why?
(c) The charged currents in the outer conducting regions of the earth's core are thought to be responsible for earth's magnetism. What might be the 'battery' (i.e., the source of energy) to sustain these currents?
(d) The earth may have even reversed the direction of its field several times during its history of 4 to 5 billion years. How can geologists know about the earth's field in such distant past?
(e) The earth's field departs from its dipole shape substantially at large distances (greater than about 30,000 km). What agencies may be responsible for this distortion?
(f) Interstellar space has an extremely weak magnetic field of the order of \(10^{-12} \, \text{T}\). Can such a weak field be of any significant consequence? Explain.
Answer & Explanation:
(a) Yes, it changes with time — secular variation over hundreds of years.
(b) Iron in core is too hot to retain permanent magnetization (above Curie temperature).
(c) Battery: Thermal convection + Coriolis effect → dynamo effect.
(d) Paleomagnetism: Magnetic alignment in ancient rocks & sediments.
(e) Distortion: Solar wind, magnetosphere, cosmic particles.
(f) Yes, affects cosmic ray paths, star formation, galactic structure.
(b) Iron in core is too hot to retain permanent magnetization (above Curie temperature).
(c) Battery: Thermal convection + Coriolis effect → dynamo effect.
(d) Paleomagnetism: Magnetic alignment in ancient rocks & sediments.
(e) Distortion: Solar wind, magnetosphere, cosmic particles.
(f) Yes, affects cosmic ray paths, star formation, galactic structure.
Question 5.3
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 \(4.5 \times 10^{-2} \, \text{J}\). What is the magnitude of magnetic moment of the magnet?
Answer & Explanation:
Formula: \( \tau = m B \sin\theta \)
Calculation:
\( m = \frac{\tau}{B \sin\theta} = \frac{4.5 \times 10^{-2}}{0.25 \times \sin 30^\circ} = \frac{0.045}{0.25 \times 0.5} = 0.36 \, \text{J T}^{-1} \)
Answer: \( 0.36 \, \text{J T}^{-1} \)
Calculation:
\( m = \frac{\tau}{B \sin\theta} = \frac{4.5 \times 10^{-2}}{0.25 \times \sin 30^\circ} = \frac{0.045}{0.25 \times 0.5} = 0.36 \, \text{J T}^{-1} \)
Answer: \( 0.36 \, \text{J T}^{-1} \)
Question 5.4
A short bar magnet of magnetic moment \( m = 0.32 \, \text{J T}^{-1} \) 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?
Answer & Explanation:
Potential energy: \( U = -mB \cos\theta \)
(a) Stable: \( \theta = 0^\circ \) (m parallel to B)
\( U_{\text{stable}} = -0.32 \times 0.15 = -0.048 \, \text{J} \)
(b) Unstable: \( \theta = 180^\circ \) (m opposite to B)
\( U_{\text{unstable}} = 0.32 \times 0.15 = 0.048 \, \text{J} \)
Answer: Stable at θ=0°, U=-0.048 J; Unstable at θ=180°, U=+0.048 J.
(a) Stable: \( \theta = 0^\circ \) (m parallel to B)
\( U_{\text{stable}} = -0.32 \times 0.15 = -0.048 \, \text{J} \)
(b) Unstable: \( \theta = 180^\circ \) (m opposite to B)
\( U_{\text{unstable}} = 0.32 \times 0.15 = 0.048 \, \text{J} \)
Answer: Stable at θ=0°, U=-0.048 J; Unstable at θ=180°, U=+0.048 J.
Question 5.5
A closely wound solenoid of 800 turns and area of cross section \(2.5 \times 10^{-4} \, \text{m}^2\) 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?
Answer & Explanation:
Acts like bar magnet because it has a north and south pole, and produces a dipole field.
Magnetic moment: \( m = N I A \)
\( m = 800 \times 3.0 \times 2.5 \times 10^{-4} = 0.6 \, \text{J T}^{-1} \)
Answer: \( 0.6 \, \text{J T}^{-1} \)
Magnetic moment: \( m = N I A \)
\( m = 800 \times 3.0 \times 2.5 \times 10^{-4} = 0.6 \, \text{J T}^{-1} \)
Answer: \( 0.6 \, \text{J T}^{-1} \)
Question 5.6
If the solenoid in Exercise 5.5 is free to turn about the vertical direction and a uniform horizontal magnetic field of 0.25 T is applied, what is the magnitude of torque on the solenoid when its axis makes an angle of \(30^\circ\) with the direction of applied field?
Answer & Explanation:
Torque: \( \tau = m B \sin\theta \)
From 5.5, \( m = 0.6 \, \text{J T}^{-1} \)
Calculation:
\( \tau = 0.6 \times 0.25 \times \sin 30^\circ = 0.6 \times 0.25 \times 0.5 = 0.075 \, \text{N·m} \)
Answer: \( 0.075 \, \text{N·m} \)
From 5.5, \( m = 0.6 \, \text{J T}^{-1} \)
Calculation:
\( \tau = 0.6 \times 0.25 \times \sin 30^\circ = 0.6 \times 0.25 \times 0.5 = 0.075 \, \text{N·m} \)
Answer: \( 0.075 \, \text{N·m} \)
Question 5.7
A bar magnet of magnetic moment \(1.5 \, \text{J T}^{-1}\) 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)?
Answer & Explanation:
Work done: \( W = mB(\cos\theta_1 - \cos\theta_2) \)
Initially θ₁ = 0°, mB = \(1.5 \times 0.22 = 0.33 \, \text{J}\)
(a)(i) θ₂ = 90°: \( W = 0.33 (\cos 0° - \cos 90°) = 0.33(1 - 0) = 0.33 \, \text{J} \)
(ii) θ₂ = 180°: \( W = 0.33 (\cos 0° - \cos 180°) = 0.33(1 - (-1)) = 0.66 \, \text{J} \)
(b) Torque: \( \tau = mB \sin\theta \)
(i) θ = 90°: \( \tau = 0.33 \times 1 = 0.33 \, \text{N·m} \)
(ii) θ = 180°: \( \tau = 0 \)
Answer: (a)(i) 0.33 J, (ii) 0.66 J; (b)(i) 0.33 N·m, (ii) 0.
Initially θ₁ = 0°, mB = \(1.5 \times 0.22 = 0.33 \, \text{J}\)
(a)(i) θ₂ = 90°: \( W = 0.33 (\cos 0° - \cos 90°) = 0.33(1 - 0) = 0.33 \, \text{J} \)
(ii) θ₂ = 180°: \( W = 0.33 (\cos 0° - \cos 180°) = 0.33(1 - (-1)) = 0.66 \, \text{J} \)
(b) Torque: \( \tau = mB \sin\theta \)
(i) θ = 90°: \( \tau = 0.33 \times 1 = 0.33 \, \text{N·m} \)
(ii) θ = 180°: \( \tau = 0 \)
Answer: (a)(i) 0.33 J, (ii) 0.66 J; (b)(i) 0.33 N·m, (ii) 0.
Question 5.8
A closely wound solenoid of 2000 turns and area of cross-section \(1.6 \times 10^{-4} \, \text{m}^2\), 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 \(7.5 \times 10^{-2} \, \text{T}\) is set up at an angle of \(30^\circ\) with the axis of the solenoid?
Answer & Explanation:
(a) Magnetic moment: \( m = NIA = 2000 \times 4.0 \times 1.6 \times 10^{-4} = 1.28 \, \text{J T}^{-1} \)
(b) Force on solenoid in uniform field: Net force = 0.
Torque: \( \tau = mB \sin\theta = 1.28 \times 7.5 \times 10^{-2} \times \sin 30^\circ \)
\( \tau = 1.28 \times 0.075 \times 0.5 = 0.048 \, \text{N·m} \)
Answer: (a) 1.28 J T⁻¹, (b) Force = 0, Torque = 0.048 N·m.
(b) Force on solenoid in uniform field: Net force = 0.
Torque: \( \tau = mB \sin\theta = 1.28 \times 7.5 \times 10^{-2} \times \sin 30^\circ \)
\( \tau = 1.28 \times 0.075 \times 0.5 = 0.048 \, \text{N·m} \)
Answer: (a) 1.28 J T⁻¹, (b) Force = 0, Torque = 0.048 N·m.
Question 5.9
A circular coil of 16 turns and radius 10 cm carrying a current of 0.75 A rests with its plane normal to an external field of magnitude \(5.0 \times 10^{-2} \, \text{T}\). The coil is free to turn about an axis in its plane perpendicular to the field direction. When the coil is turned slightly and released, it oscillates about its stable equilibrium with a frequency of 2.0 s⁻¹. What is the moment of inertia of the coil about its axis of rotation?
Answer & Explanation:
Formula for oscillating magnetic dipole: \( T = 2\pi \sqrt{\frac{I}{mB}} \)
Frequency \( f = 2 \, \text{s}^{-1} \) → \( T = 0.5 \, \text{s} \)
Magnetic moment \( m = NIA = 16 \times 0.75 \times \pi (0.1)^2 = 0.3768 \, \text{J T}^{-1} \)
Calculation:
\( I = \frac{T^2 mB}{4\pi^2} = \frac{(0.5)^2 \times 0.3768 \times 5.0 \times 10^{-2}}{4\pi^2} \approx 1.19 \times 10^{-4} \, \text{kg m}^2 \)
Answer: \( 1.19 \times 10^{-4} \, \text{kg m}^2 \)
Frequency \( f = 2 \, \text{s}^{-1} \) → \( T = 0.5 \, \text{s} \)
Magnetic moment \( m = NIA = 16 \times 0.75 \times \pi (0.1)^2 = 0.3768 \, \text{J T}^{-1} \)
Calculation:
\( I = \frac{T^2 mB}{4\pi^2} = \frac{(0.5)^2 \times 0.3768 \times 5.0 \times 10^{-2}}{4\pi^2} \approx 1.19 \times 10^{-4} \, \text{kg m}^2 \)
Answer: \( 1.19 \times 10^{-4} \, \text{kg m}^2 \)
Question 5.10
A magnetic needle free to rotate in a vertical plane parallel to the magnetic meridian has its north tip pointing down at \(22^\circ\) with the horizontal. The horizontal component of the earth's magnetic field at the place is known to be 0.35 G. Determine the magnitude of the earth's magnetic field at the place.
Answer & Explanation:
Given: Angle of dip δ = 22°, \( B_H = 0.35 \, \text{G} = 0.35 \times 10^{-4} \, \text{T} \)
Relation: \( B_H = B \cos\delta \) → \( B = \frac{B_H}{\cos\delta} \)
Calculation:
\( B = \frac{0.35 \times 10^{-4}}{\cos 22^\circ} = \frac{3.5 \times 10^{-5}}{0.9272} \approx 3.77 \times 10^{-5} \, \text{T} = 0.377 \, \text{G} \)
Answer: \( 0.377 \, \text{G} \)
Relation: \( B_H = B \cos\delta \) → \( B = \frac{B_H}{\cos\delta} \)
Calculation:
\( B = \frac{0.35 \times 10^{-4}}{\cos 22^\circ} = \frac{3.5 \times 10^{-5}}{0.9272} \approx 3.77 \times 10^{-5} \, \text{T} = 0.377 \, \text{G} \)
Answer: \( 0.377 \, \text{G} \)
Question 5.11
At a certain location in Africa, a compass points \(12^\circ\) west of the geographic north. The north tip of the magnetic needle of a dip circle placed in the plane of magnetic meridian points \(60^\circ\) above the horizontal. The horizontal component of the earth's field is measured to be 0.16 G. Specify the direction and magnitude of the earth's field at the location.
Answer & Explanation:
Given: Declination = 12° W, Dip δ = 60°, \( B_H = 0.16 \, \text{G} \)
Magnitude: \( B = \frac{B_H}{\cos\delta} = \frac{0.16}{\cos 60^\circ} = \frac{0.16}{0.5} = 0.32 \, \text{G} \)
Direction: Magnetic field direction is at 60° to horizontal in magnetic meridian plane, which is 12° west of geographic north.
Answer: 0.32 G, inclined at 60° to horizontal in plane 12° west of geographic north.
Magnitude: \( B = \frac{B_H}{\cos\delta} = \frac{0.16}{\cos 60^\circ} = \frac{0.16}{0.5} = 0.32 \, \text{G} \)
Direction: Magnetic field direction is at 60° to horizontal in magnetic meridian plane, which is 12° west of geographic north.
Answer: 0.32 G, inclined at 60° to horizontal in plane 12° west of geographic north.
Question 5.12
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.
Answer & Explanation:
Axial field: \( B_{\text{axis}} = \frac{\mu_0}{4\pi} \frac{2m}{r^3} \)
Equatorial field: \( B_{\text{eq}} = \frac{\mu_0}{4\pi} \frac{m}{r^3} \)
Calculation:
\( r = 0.1 \, \text{m}, \, \frac{\mu_0}{4\pi} = 10^{-7} \)
\( B_{\text{axis}} = 10^{-7} \times \frac{2 \times 0.48}{(0.1)^3} = 9.6 \times 10^{-5} \, \text{T} \) (along axis, N to S)
\( B_{\text{eq}} = 10^{-7} \times \frac{0.48}{(0.1)^3} = 4.8 \times 10^{-5} \, \text{T} \) (opposite to m direction)
Answer: (a) \( 9.6 \times 10^{-5} \, \text{T} \) along axis, (b) \( 4.8 \times 10^{-5} \, \text{T} \) opposite to m.
Equatorial field: \( B_{\text{eq}} = \frac{\mu_0}{4\pi} \frac{m}{r^3} \)
Calculation:
\( r = 0.1 \, \text{m}, \, \frac{\mu_0}{4\pi} = 10^{-7} \)
\( B_{\text{axis}} = 10^{-7} \times \frac{2 \times 0.48}{(0.1)^3} = 9.6 \times 10^{-5} \, \text{T} \) (along axis, N to S)
\( B_{\text{eq}} = 10^{-7} \times \frac{0.48}{(0.1)^3} = 4.8 \times 10^{-5} \, \text{T} \) (opposite to m direction)
Answer: (a) \( 9.6 \times 10^{-5} \, \text{T} \) along axis, (b) \( 4.8 \times 10^{-5} \, \text{T} \) opposite to m.
Question 5.13
A short bar magnet placed in a horizontal plane has its axis aligned along the magnetic north‑south direction. Null points are found on the axis of the magnet at 14 cm from the centre of the magnet. The earth's magnetic field at the place is 0.36 G and the angle of dip is zero. What is the total magnetic field on the normal bisector of the magnet at the same distance as the null‑point (i.e., 14 cm) from the centre of the magnet? (At null points, field due to a magnet is equal and opposite to the horizontal component of earth's magnetic field.)
Answer & Explanation:
At null point on axis: \( B_{\text{axis}} = B_H \) (earth's horizontal field)
\( \frac{\mu_0}{4\pi} \frac{2m}{r^3} = 0.36 \times 10^{-4} \, \text{T} \)
On equatorial line: \( B_{\text{eq}} = \frac{\mu_0}{4\pi} \frac{m}{r^3} = \frac{B_H}{2} = 0.18 \times 10^{-4} \, \text{T} \)
Total field = vector sum of \( B_{\text{eq}} \) and \( B_H \), perpendicular to each other.
\( B_{\text{total}} = \sqrt{B_H^2 + B_{\text{eq}}^2} = \sqrt{(0.36)^2 + (0.18)^2} \times 10^{-4} \, \text{T} \)
\( \approx 0.402 \times 10^{-4} \, \text{T} = 0.402 \, \text{G} \)
Answer: \( 0.402 \, \text{G} \)
\( \frac{\mu_0}{4\pi} \frac{2m}{r^3} = 0.36 \times 10^{-4} \, \text{T} \)
On equatorial line: \( B_{\text{eq}} = \frac{\mu_0}{4\pi} \frac{m}{r^3} = \frac{B_H}{2} = 0.18 \times 10^{-4} \, \text{T} \)
Total field = vector sum of \( B_{\text{eq}} \) and \( B_H \), perpendicular to each other.
\( B_{\text{total}} = \sqrt{B_H^2 + B_{\text{eq}}^2} = \sqrt{(0.36)^2 + (0.18)^2} \times 10^{-4} \, \text{T} \)
\( \approx 0.402 \times 10^{-4} \, \text{T} = 0.402 \, \text{G} \)
Answer: \( 0.402 \, \text{G} \)
Question 5.14
If the bar magnet in exercise 5.13 is turned around by \(180^\circ\), where will the new null points be located?
Answer & Explanation:
Turning magnet by 180° reverses its field direction. Null points occur where magnet's field cancels earth's horizontal field.
Since earth's field direction unchanged, null points will now be on equatorial line (perpendicular to axis) at same distance r = 14 cm.
Answer: On equatorial line at 14 cm from centre.
Since earth's field direction unchanged, null points will now be on equatorial line (perpendicular to axis) at same distance r = 14 cm.
Answer: On equatorial line at 14 cm from centre.
Question 5.15
A short bar magnet of magnetic moment \( 5.25 \times 10^{-2} \, \text{J T}^{-1} \) is placed with its axis perpendicular to the earth's field direction. At what distance from the centre of the magnet, the resultant field is inclined at \( 45^\circ \) with earth's field on (a) its normal bisector and (b) its axis. Magnitude of the earth's field at the place is given to be \( 0.42 \, \text{G} \). Ignore the length of the magnet in comparison to the distances involved.
Answer & Explanation:
Earth's field \( B_E = 0.42 \times 10^{-4} \, \text{T} \)
(a) On equatorial line: Magnet field \( B_m = \frac{\mu_0 m}{4\pi r^3} \) ⊥ to \( B_E \)
For 45° inclination: \( \tan 45^\circ = 1 = \frac{B_m}{B_E} \) → \( B_m = B_E \)
\( \frac{10^{-7} \times 5.25 \times 10^{-2}}{r^3} = 0.42 \times 10^{-4} \)
\( r^3 = \frac{5.25 \times 10^{-9}}{0.42 \times 10^{-4}} = 1.25 \times 10^{-4} \)
\( r \approx 0.05 \, \text{m} = 5.0 \, \text{cm} \)
(b) On axis: \( B_m = \frac{\mu_0}{4\pi} \frac{2m}{r^3} \) parallel/antiparallel to \( B_E \)
For 45°: \( B_m = B_E \) (same magnitude, different direction)
\( \frac{10^{-7} \times 2 \times 5.25 \times 10^{-2}}{r^3} = 0.42 \times 10^{-4} \)
\( r^3 = \frac{1.05 \times 10^{-8}}{0.42 \times 10^{-4}} = 2.5 \times 10^{-4} \)
\( r \approx 0.063 \, \text{m} = 6.3 \, \text{cm} \)
Answer: (a) 5.0 cm, (b) 6.3 cm.
(a) On equatorial line: Magnet field \( B_m = \frac{\mu_0 m}{4\pi r^3} \) ⊥ to \( B_E \)
For 45° inclination: \( \tan 45^\circ = 1 = \frac{B_m}{B_E} \) → \( B_m = B_E \)
\( \frac{10^{-7} \times 5.25 \times 10^{-2}}{r^3} = 0.42 \times 10^{-4} \)
\( r^3 = \frac{5.25 \times 10^{-9}}{0.42 \times 10^{-4}} = 1.25 \times 10^{-4} \)
\( r \approx 0.05 \, \text{m} = 5.0 \, \text{cm} \)
(b) On axis: \( B_m = \frac{\mu_0}{4\pi} \frac{2m}{r^3} \) parallel/antiparallel to \( B_E \)
For 45°: \( B_m = B_E \) (same magnitude, different direction)
\( \frac{10^{-7} \times 2 \times 5.25 \times 10^{-2}}{r^3} = 0.42 \times 10^{-4} \)
\( r^3 = \frac{1.05 \times 10^{-8}}{0.42 \times 10^{-4}} = 2.5 \times 10^{-4} \)
\( r \approx 0.063 \, \text{m} = 6.3 \, \text{cm} \)
Answer: (a) 5.0 cm, (b) 6.3 cm.
Question 5.16
Answer the following questions:
(a) Why does a paramagnetic sample display greater magnetisation (for the same magnetising field) when cooled?
(b) Why is diamagnetism, in contrast, almost independent of temperature?
(c) If a toroid uses bismuth for its core, will the field in the core be (slightly) greater or (slightly) less than when the core is empty?
(d) Is the permeability of a ferromagnetic material independent of the magnetic field? If not, is it more for lower or higher fields?
(e) Magnetic field lines are always nearly normal to the surface of a ferromagnet at every point. (This fact is analogous to the static electric field lines being normal to the surface of a conductor at every point.) Why?
(f) Would the maximum possible magnetisation of a paramagnetic sample be of the same order of magnitude as the magnetisation of a ferromagnet?
Answer & Explanation:
(a) Cooling reduces thermal agitation, allowing better alignment of magnetic dipoles.
(b) Diamagnetism arises from induced orbital motion, independent of thermal motion.
(c) Slightly less, bismuth is diamagnetic (μᵣ < 1).
(d) Not independent; permeability decreases with increasing field (saturation).
(e) Because tangential component of B must be continuous, and inside ferromagnet B is large → outside tangential component negligible.
(f) No, ferromagnets have orders of magnitude higher magnetisation (domain alignment).
(b) Diamagnetism arises from induced orbital motion, independent of thermal motion.
(c) Slightly less, bismuth is diamagnetic (μᵣ < 1).
(d) Not independent; permeability decreases with increasing field (saturation).
(e) Because tangential component of B must be continuous, and inside ferromagnet B is large → outside tangential component negligible.
(f) No, ferromagnets have orders of magnitude higher magnetisation (domain alignment).
Question 5.17
Answer the following questions:
(a) Explain qualitatively on the basis of domain picture the irreversibility in the magnetisation curve of a ferromagnet.
(b) The hysteresis loop of a soft iron piece has a much smaller area than that of a carbon steel piece. If the material is to go through repeated cycles of magnetisation, which piece will dissipate greater heat energy?
(c) 'A system displaying a hysteresis loop such as a ferromagnet, is a device for storing memory?' Explain the meaning of this statement.
(d) What kind of ferromagnetic material is used for coating magnetic tapes in a cassette player, or for building 'memory stores' in a modern computer?
(e) A certain region of space is to be shielded from magnetic fields. Suggest a method.
Answer & Explanation:
(a) Domain walls do not retrace path due to pinning, energy loss → hysteresis.
(b) Carbon steel (larger loop area → greater energy loss per cycle).
(c) Hysteresis allows retention of magnetisation after field removal → binary memory storage.
(d) Hard ferromagnetic materials (high coercivity) e.g., CrO₂, ferrites.
(e) Use high‑μ material enclosure (magnetic shielding) e.g., soft iron box.
(b) Carbon steel (larger loop area → greater energy loss per cycle).
(c) Hysteresis allows retention of magnetisation after field removal → binary memory storage.
(d) Hard ferromagnetic materials (high coercivity) e.g., CrO₂, ferrites.
(e) Use high‑μ material enclosure (magnetic shielding) e.g., soft iron box.
Question 5.18
A long straight horizontal cable carries a current of \( 2.5 \, \text{A} \) in the direction \( 10^\circ \) south of west to \( 10^\circ \) north of east. The magnetic meridian of the place happens to be \( 10^\circ \) west of the geographic meridian. The earth's magnetic field at the location is \( 0.33 \, \text{G} \), and the angle of dip is zero. Locate the line of neutral points (ignore the thickness of the cable). (At neutral points, magnetic field due to a current‑carrying cable is equal and opposite to the horizontal component of earth's magnetic field.)
Answer & Explanation:
Cable direction: 10° S of W to 10° N of E → essentially E‑W with slight tilt.
Earth's horizontal field \( B_H = 0.33 \, \text{G} \) in magnetic meridian (10° W of N).
Neutral points where \( B_{\text{wire}} = B_H \).
\( B_{\text{wire}} = \frac{\mu_0 I}{2\pi r} \)
\( r = \frac{\mu_0 I}{2\pi B_H} = \frac{4\pi \times 10^{-7} \times 2.5}{2\pi \times 0.33 \times 10^{-4}} \approx 1.52 \times 10^{-2} \, \text{m} = 1.52 \, \text{cm} \)
Neutral points lie on line perpendicular to cable at this distance.
Answer: Line parallel to cable, 1.52 cm above/below it.
Earth's horizontal field \( B_H = 0.33 \, \text{G} \) in magnetic meridian (10° W of N).
Neutral points where \( B_{\text{wire}} = B_H \).
\( B_{\text{wire}} = \frac{\mu_0 I}{2\pi r} \)
\( r = \frac{\mu_0 I}{2\pi B_H} = \frac{4\pi \times 10^{-7} \times 2.5}{2\pi \times 0.33 \times 10^{-4}} \approx 1.52 \times 10^{-2} \, \text{m} = 1.52 \, \text{cm} \)
Neutral points lie on line perpendicular to cable at this distance.
Answer: Line parallel to cable, 1.52 cm above/below it.
Question 5.19
A telephone cable at a place has four long straight horizontal wires carrying a current of \( 1.0 \, \text{A} \) in the same direction east to west. The earth's magnetic field at the place is 0.39 G, and the angle of dip is 35°. The magnetic declination is nearly zero. What are the resultant magnetic fields at points 4.0 cm below the cable?
Answer & Explanation:
Earth's field: \( B_E = 0.39 \, \text{G} \), dip = 35° → \( B_H = B_E \cos 35^\circ \), \( B_V = B_E \sin 35^\circ \).
Wires produce magnetic field at 4 cm below: \( B_{\text{wire}} = \frac{\mu_0 \times 4I}{2\pi r} \) (4 wires same direction)
\( B_{\text{wire}} = \frac{4\pi \times 10^{-7} \times 4 \times 1.0}{2\pi \times 0.04} = 2.0 \times 10^{-5} \, \text{T} = 0.2 \, \text{G} \)
Direction: By right‑hand rule, below east‑west wire, field is north‑south.
Resultant = vector sum of \( B_{\text{wire}} \), \( B_H \), \( B_V \).
Detailed calculation: Resolve components and add.
Answer: ≈ 0.49 G total field (direction depends on exact geometry).
Wires produce magnetic field at 4 cm below: \( B_{\text{wire}} = \frac{\mu_0 \times 4I}{2\pi r} \) (4 wires same direction)
\( B_{\text{wire}} = \frac{4\pi \times 10^{-7} \times 4 \times 1.0}{2\pi \times 0.04} = 2.0 \times 10^{-5} \, \text{T} = 0.2 \, \text{G} \)
Direction: By right‑hand rule, below east‑west wire, field is north‑south.
Resultant = vector sum of \( B_{\text{wire}} \), \( B_H \), \( B_V \).
Detailed calculation: Resolve components and add.
Answer: ≈ 0.49 G total field (direction depends on exact geometry).
Question 5.20
A compass needle free to turn in a horizontal plane is placed at the centre of circular coil of 30 turns and radius 12 cm. The coil is in a vertical plane making an angle of 45° with the magnetic meridian. When the current in the coil is 0.35 A, the needle points west to east.
(a) Determine the horizontal component of the earth's magnetic field at the location.
(b) The current in the coil is reversed, and the coil is rotated about its vertical axis by an angle of 90° in the anticlockwise sense looking from above. Predict the direction of the needle. Take the magnetic declination at the places to be zero.
Answer & Explanation:
(a) Coil field at centre: \( B_c = \frac{\mu_0 NI}{2R} \)
\( B_c = \frac{4\pi \times 10^{-7} \times 30 \times 0.35}{2 \times 0.12} \approx 5.5 \times 10^{-5} \, \text{T} \)
Needle points W‑E → coil field cancels earth's N‑S component.
\( B_c \cos 45^\circ = B_H \) → \( B_H = B_c \cos 45^\circ = 3.89 \times 10^{-5} \, \text{T} \)
(b) Reversing current flips coil field direction. Rotating coil 90° changes its field orientation relative to earth's field.
Needle will align with net horizontal field.
Answer: (a) \( 3.89 \times 10^{-5} \, \text{T} \), (b) Needle turns accordingly (specific direction needs diagram).
\( B_c = \frac{4\pi \times 10^{-7} \times 30 \times 0.35}{2 \times 0.12} \approx 5.5 \times 10^{-5} \, \text{T} \)
Needle points W‑E → coil field cancels earth's N‑S component.
\( B_c \cos 45^\circ = B_H \) → \( B_H = B_c \cos 45^\circ = 3.89 \times 10^{-5} \, \text{T} \)
(b) Reversing current flips coil field direction. Rotating coil 90° changes its field orientation relative to earth's field.
Needle will align with net horizontal field.
Answer: (a) \( 3.89 \times 10^{-5} \, \text{T} \), (b) Needle turns accordingly (specific direction needs diagram).
Question 5.21
A magnetic dipole is under the influence of two magnetic fields. The angle between the field directions is 60°, and one of the fields has a magnitude of \(1.2 \times 10^{-2} \, \text{T}\). If the dipole comes to stable equilibrium at an angle of 15° with this field, what is the magnitude of the other field?
Answer & Explanation:
Let fields be \( \vec{B_1} \) and \( \vec{B_2} \), angle between them = 60°.
Dipole makes 15° with \( B_1 \) at equilibrium.
Torque balance: \( m B_1 \sin 15^\circ = m B_2 \sin(60^\circ - 15^\circ) \)
\( B_2 = B_1 \frac{\sin 15^\circ}{\sin 45^\circ} = 1.2 \times 10^{-2} \times \frac{0.2588}{0.7071} \approx 4.39 \times 10^{-3} \, \text{T} \)
Answer: \( 4.39 \times 10^{-3} \, \text{T} \)
Dipole makes 15° with \( B_1 \) at equilibrium.
Torque balance: \( m B_1 \sin 15^\circ = m B_2 \sin(60^\circ - 15^\circ) \)
\( B_2 = B_1 \frac{\sin 15^\circ}{\sin 45^\circ} = 1.2 \times 10^{-2} \times \frac{0.2588}{0.7071} \approx 4.39 \times 10^{-3} \, \text{T} \)
Answer: \( 4.39 \times 10^{-3} \, \text{T} \)
Question 5.22
A monoenergetic (18 keV) electron beam initially in the horizontal direction is subjected to a horizontal magnetic field of 0.04 G normal to the initial direction. Estimate the up or down deflection of the beam over a distance of 30 cm (\( m_e = 9.11 \times 10^{-31} \, \text{kg}, \, e = 1.6 \times 10^{-19} \, \text{C} \)). [Note: Data in this exercise are so chosen that the answer will give you an idea of the effect of earth's magnetic field on the motion of the electron beam from the electron gun to the screen in a TV set.]
Answer & Explanation:
Kinetic energy = 18 keV = \( 18 \times 10^3 \times 1.6 \times 10^{-19} \, \text{J} \)
\( v = \sqrt{\frac{2K}{m}} \) ≈ \( 2.5 \times 10^7 \, \text{m/s} \)
Magnetic field \( B = 0.04 \, \text{G} = 4 \times 10^{-6} \, \text{T} \)
Radius of curvature: \( r = \frac{mv}{eB} \) ≈ 35.5 m
Over small arc length L = 0.3 m, deflection \( y \approx \frac{L^2}{2r} \) ≈ 1.27 × 10⁻³ m ≈ 1.3 mm
Answer: ~1.3 mm deflection.
\( v = \sqrt{\frac{2K}{m}} \) ≈ \( 2.5 \times 10^7 \, \text{m/s} \)
Magnetic field \( B = 0.04 \, \text{G} = 4 \times 10^{-6} \, \text{T} \)
Radius of curvature: \( r = \frac{mv}{eB} \) ≈ 35.5 m
Over small arc length L = 0.3 m, deflection \( y \approx \frac{L^2}{2r} \) ≈ 1.27 × 10⁻³ m ≈ 1.3 mm
Answer: ~1.3 mm deflection.
Question 5.23
A sample of paramagnetic salt contains \(2.0 \times 10^{24}\) atomic dipoles each of dipole moment \(1.5 \times 10^{-23} \, \text{J T}^{-1}\). The sample is placed under a homogeneous magnetic field of 0.64 T, and cooled to a temperature of 4.2 K. The degree of magnetic saturation achieved is equal to 15%. What is the total dipole moment of the sample for a magnetic field of 0.98 T and a temperature of 2.8 K? (Assume Curie's law)
Answer & Explanation:
Curie's law: Magnetisation \( M \propto \frac{B}{T} \)
Saturation magnetisation \( M_s = N \mu \) = total possible alignment.
Given 15% saturation at B₁=0.64 T, T₁=4.2 K.
For new B₂=0.98 T, T₂=2.8 K:
\( \frac{M_2}{M_1} = \frac{B_2/T_2}{B_1/T_1} \) → \( M_2 = 0.15 \times \frac{0.98/2.8}{0.64/4.2} \times (N\mu) \)
Total dipole moment = \( M_2 \times \text{volume} \) — but here Nμ known directly.
\( N\mu = 2.0 \times 10^{24} \times 1.5 \times 10^{-23} = 30 \, \text{J T}^{-1} \)
\( M_2 = 0.15 \times \frac{0.98/2.8}{0.64/4.2} \times 30 \approx 0.15 \times 2.296 \times 30 \approx 10.33 \, \text{J T}^{-1} \)
Answer: ~10.33 J T⁻¹
Saturation magnetisation \( M_s = N \mu \) = total possible alignment.
Given 15% saturation at B₁=0.64 T, T₁=4.2 K.
For new B₂=0.98 T, T₂=2.8 K:
\( \frac{M_2}{M_1} = \frac{B_2/T_2}{B_1/T_1} \) → \( M_2 = 0.15 \times \frac{0.98/2.8}{0.64/4.2} \times (N\mu) \)
Total dipole moment = \( M_2 \times \text{volume} \) — but here Nμ known directly.
\( N\mu = 2.0 \times 10^{24} \times 1.5 \times 10^{-23} = 30 \, \text{J T}^{-1} \)
\( M_2 = 0.15 \times \frac{0.98/2.8}{0.64/4.2} \times 30 \approx 0.15 \times 2.296 \times 30 \approx 10.33 \, \text{J T}^{-1} \)
Answer: ~10.33 J T⁻¹
Question 5.24
A Rowland ring of mean radius 15 cm has 3500 turns of wire wound on a ferromagnetic core of relative permeability 800. What is the magnetic field B in the core for a magnetising current of 1.2 A?
Answer & Explanation:
Rowland ring: \( B = \mu_0 \mu_r n I \) where \( n = \frac{N}{2\pi r} \)
\( n = \frac{3500}{2\pi \times 0.15} \approx 3715 \, \text{turns/m} \)
\( B = 4\pi \times 10^{-7} \times 800 \times 3715 \times 1.2 \approx 4.48 \, \text{T} \)
Answer: ~4.48 T
\( n = \frac{3500}{2\pi \times 0.15} \approx 3715 \, \text{turns/m} \)
\( B = 4\pi \times 10^{-7} \times 800 \times 3715 \times 1.2 \approx 4.48 \, \text{T} \)
Answer: ~4.48 T
Question 5.25
The magnetic moment vectors \( \mu_s \) and \( \mu_l \) associated with the intrinsic spin angular momentum S and orbital angular momentum I, respectively, of an electron are predicted by quantum theory (and verified experimentally to a high accuracy) to be given by:
\( \mu_s = -(e/m) \, \mathbf{S} \),
\( \mu_l = -(e/2m) \, \mathbf{I} \).
Which of these relations is in accordance with the result expected classically? Outline the derivation of the classical result.
\( \mu_s = -(e/m) \, \mathbf{S} \),
\( \mu_l = -(e/2m) \, \mathbf{I} \).
Which of these relations is in accordance with the result expected classically? Outline the derivation of the classical result.
Answer & Explanation:
Classical result: Magnetic moment due to orbital motion: \( \mu_l = I A = \frac{e}{2m} L \) (L = orbital angular momentum).
This matches \( \mu_l = -(e/2m) \mathbf{I} \) (ignoring sign).
Spin magnetic moment has no classical analogue → \( \mu_s = -(e/m) \mathbf{S} \) is purely quantum.
Classical derivation: For circular orbit, current \( I = \frac{e}{T} = \frac{e \omega}{2\pi} \), area \( A = \pi r^2 \).
Angular momentum \( L = m r^2 \omega \) → \( \mu = I A = \frac{e \omega}{2\pi} \cdot \pi r^2 = \frac{e}{2m} L \).
Answer: \( \mu_l \) relation agrees with classical expectation.
This matches \( \mu_l = -(e/2m) \mathbf{I} \) (ignoring sign).
Spin magnetic moment has no classical analogue → \( \mu_s = -(e/m) \mathbf{S} \) is purely quantum.
Classical derivation: For circular orbit, current \( I = \frac{e}{T} = \frac{e \omega}{2\pi} \), area \( A = \pi r^2 \).
Angular momentum \( L = m r^2 \omega \) → \( \mu = I A = \frac{e \omega}{2\pi} \cdot \pi r^2 = \frac{e}{2m} L \).
Answer: \( \mu_l \) relation agrees with classical expectation.
📘 Exam Preparation Tip:
These exercise questions will help you understand magnetic properties of materials and geomagnetism. You'll learn to classify magnetic materials (diamagnetic, paramagnetic, ferromagnetic), analyze magnetic field lines and poles, calculate magnetic dipole moment and torque, understand Earth's magnetic field components, and interpret hysteresis curves for ferromagnetic materials. Essential for material science and geophysics applications.