NCERT Physics Solution : Magnetism and Matter

Q4. 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 ?

Sol. M= 0.32 JT–1 , B = 0.15 T

(a) by using formula , U = −MBcosθ

for stable equilibrium , θ = 0º

U = −MBcosθ = −0.32 x 0.15×1 joule

(b) for unstable equilibrium , θ = 180º

U = −MBcosθ = −0.32 x 0.15 x(−1) joule

Q5. 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 ?

Sol. N = 800 , A = 2.5 × 10–4 m2 , I = 3 A

By using formula , M = NIA

whenever the current is passed through the solenoid , the magnetic field is produced along its axis . The magnetic field lines come out from one end and enter the other end of the solenoid .The ends act as bar magnet poles . Magnetic moment , M = NIA = 800 × 3 × 2.5 × 10–4 Am2

M = 0.6 Am2

Q6. 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?

Sol.

B = 0.25 T , M = 0.6 Am2

angle , θ = 30º

By using formula , τ = MBsinθ

τ = 0.6 x 0.25 x 0.5 = 0.075 Nm

Q7. 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) ?

Sol.