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NCERT Class 12 Physics Chapter 5 Magnetism and Matter
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Magnetism and Matter
Chapter: 5
| Part – I |
EXERCISE
1. A short bar magnet placed with its axis at 30° with a uniform external magnetic field of 0.25 T experiences a torque of magnitude equal to 4.5 × 10–2 J. What is the magnitude of the magnetic moment of the magnet?
Ans: Here,
τ = μB sin θ
Given:
τ = 4.5 × 10-2 J
B = 0.25 T
θ = 30° (sin = 30° = 0.5)
τ = μB sin θ
= 4.5 × 10-2 = μ × 0.25 × 0.5
= 4.5 × 10-2 = μ × 0.125
= μ = 4.5 × 10-2 / 0.125
= 0.36 Am2
2. A short bar magnet of magnetic moment m = 0.32 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?
Ans: (a) Given:
m = 0.32 JT-1
B = 0.15
(θ = 0°)
Here:
U = – m × B × cos θ
= – m × B
= – 0.32 × 0.15
= – 0.048 J.
(b) Given:
m = 0.32 JT-1
B = 0.15
(θ = 180°)
Here:
U = – m × B × cos θ
= – m × B
= – 0.32 × 0.15 (-1)
= 0.048 J.
3. A closely wound solenoid of 800 turns and area of cross section 2.5 × 10–4 m2 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?
Ans: Given:
n = 800
l = 3.0 A
A = 2.5 × 10-4 m2
Here:
m = n × l × A
m = 800 × 3.0 × 2.5 × 10-4
= 6000 × 10-4
= 0.6 Am2.
4. 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° with the direction of applied field?
Ans: Given:
m = 0.6 Am2
B = 0.25 T
τ = m × B × sin θ
τ = 0.6 × 0.25 × Sin 30° (Sin 30° = 0.5)
= 0.6 × 0.25 × 0.5
= 0.6 × 0.125
= 0.075 Nm.
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.
Ans: Given:
m = 1.5 J/T
B = 0.22 T.
W = mB (cosθ1 – cos θ2)
θ1 = 0°
θ2 = 90°
cos θ1 = cos 0° = 1
cos θ2 = cos 90° = 0
W = mB (1 – 0)
W = 1.5 × 0.22 W
= 0.33 J.
(ii) opposite to the field direction?
Ans: Given:
m = 1.5 J/T
B = 0.22 T.
W = mB (cosθ1 – cos θ2)
θ1 = 0°
θ2 = 180°
cos θ1 = cos 0° = 1
cos θ2 = cos 90° = -1
W = mB [1- (-1)]
W = 1.5 × 0.22 × 2W
= 0.66 J.
(b) What is the torque on the magnet in cases (i) and (ii)?
Ans: (i) Given:
m = 1.5 J/T
B = 0.22 T.
θ = 90° (so, sin 90° = 1)
Therefore:
τ = m × B × 1
= 1.5 × 0.22
= 0.33 Nm.
(ii) Given:
m = 1.5 J/T
B = 0.22 T.
θ = 180° (so, sin 90° = 0)
Therefore:
τ = m × B × 0
= 1.5 × 0.22 × 0
= 0 Nm.
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?
Ans: Given:
n = 2000
A = 1.6 × 10-4 m2
l = 4.0 A
m = n × l × A
= 2000 × 4.0 × 1.6 × 10-4 Am2
= 1280 × 10-4 Am2
= 0.128 Am2
(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?
Ans: B = 7.5 × 10-2 T
Angle θ = 30°
τ = m × B sin θ
= 0.128 × 7.5 × 10-2 × Sin 30°
= 0.128 × 7.5 × 10-2 × ½
= 0.128 × 3.75 × 10-2 Nm
= 0.48 × 10-2 Nm
= 4.8 × 10-3 Nm
= 0.0048 Nm.
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.
Ans: (a) On the axis:
B = μ₀ × (2m / (4𝜋d3)
μ₀ = (4𝜋 × 10-7)
M = (0.48 JT-1)
d (10 cm = 0.1 m).
B = (4𝜋 × 10-7) × (2 × 0.48 JT-1) / (4𝜋 × (0.1 m3)
= 0.096 T
(b) On the equatorial lines:
B = μ₀ × (M / (4𝜋d3)
B = (4𝜋 × 10-7) × (0.48 JT-1) / (4𝜋 × (0.1 m3)
= 0.048 T
