NCERT Solutions for Class 12 Physics Chapter 5 Magnetism and Matter - Param Himalaya
5.1 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 4.5x 10 J. What is the magnitude of magnetic moment of the magnet?
Solution:
Magnetic field strength , B = 0.25T , Torque on the bar magnet , $T = 4.5 \times 10^{-2}J$ ,
Angle between the bar magnet and the external magnetic field , $\theta = 30^{\circ}$.
then Torque is related to magnetic moment (M) as :
$$\tau = MBsin\theta$$
$$\therefore M = \frac{\tau}{Bsin\theta}$$
$$M = \frac{4.5 \times 10^{-2}}{0.25 \times sin30^{0}}$$
$$M = 0.36 JT^{-1}$$
hence , the magnetic moment of the magnet is $M = 0.36 JT^{-1}$
5.2 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?
Solution:
Potential Energy $= -MBcos\theta$
(a) If $\theta = 0^{\circ},$
Potential energy$ = -MBcos0^{\circ}$
$U = -MB$
$U= -0.32 \times 0.15$
$U= -0.048 J$
the dipole will be in stable equilibrium
(b) if $\theta = 180^{\circ},$
$U = -MBcos180^{\circ}$
$U = MB = 0.32 \times 0.15$
$U = 0.048J$
The dipole will be in unstable equilibrium.
5.3 A closely wound solenoid of 800 turns and area of cross section $2.5 \times 10^{-4}$ 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?
Solution:
Using , M = NIA , we get
$M = 800 \times 3 \times2.5\times10^{-4}$
$M = 0.60 JT^{-1}$
A current-carrying solenoid behaves as a bar magnet because a magnetic field develops along its axis, i.e., along its length.
5.4 If the solenoid in Exercise 5.3 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?
Solution:
Using $\tau = MBsin\theta$ , we get
$$\tau = 0.6 \times 0.25 \times sin 30^{0}$$
$$\tau =0.6 \times0.25 \times \frac{1}{2}$$
$$=0.3 \times 0.25= 0.075 Nm$$
$$\tau = 7.5 \times 10^{-2}Nm$$
$$\tau=0.075Nm$$
5.5 A bar magnet of magnetic moment $1.5 JT^{-1}$ lies aligned with the direction of a uniform magnetic field of 0.22T.
(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)?
Solution:
Here , M = 1.5 J /T ; B = 0.22 T
5.6 A closely wound solenoid of 2000 turns and area of cross-section $1.6 \times 10^{-4}$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 $7.5 \times 10^{2} T$ is set up at an angle of 30° with the axis of the solenoid?
Solution:
Here , n= 2000 ; $A= 1.6 \times 10^{-4}m^{2}$ ; I=4A
(a) Magnetic moment of the solenoid is
$M= nIA $
$M= 2000 \times 4 \times 1.6 \times 10^{-4}$
$M= 1.28 J/T$
(b) Net force on the solenoid (magnetic dipole) is zero in a uniform magnetic field.
Torque ,
$$\tau=MBsin\theta$$
$$\tau = 1.28 \times 7.5 \times 10^{-2} sin30^{0}$$
$$\tau = 0.048 Nm$$
5.7 A short bar magnet has a magnetic moment of 0.48 J /T. 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.
Solution:
5.8 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.)
Solution :
As the null point lie on the axis of the magnet,
$$B_{axial} = \frac{ \mu_{0}}{4 \pi}. \frac{2M}{d^{3}} = H$$
At the same distance (d) on the equatorial line of the magnet , the magnetic field due to short bar magnet is
$$B_{eqa} =\frac{ \mu_{0}}{4 \pi}. \frac{M}{d^{3}} = \frac{H}{2}$$
Total magnetic field at the required point on the equatorial line of the magnet is
$$B = B_{eqa} + H$$
$$B = \frac{H}{2} + H$$
$$B = \frac{3H}{2}$$
$$B = \frac{3 \times 0.36}{2}$$
$$B = 0.54 G$$
5.9 If the bar magnet in exercise 5.8 is turned around by 180°, where will the new null points be located?
Solution:
In this case , neutral points will lie on the equatorial line of the magnet , say at a distance x from the centre of the magnet. Then
$$B_{eqa} = \frac{\mu_{0} }{4 \pi}.\frac{M}{x^{3}} = H$$
$$H= \frac{\mu_{0} }{4 \pi}.\frac{2M}{d^{3}}$$
$$\frac{\mu_{0} }{4 \pi}.\frac{M}{x^{3}} = \frac{\mu_{0} }{4 \pi}.\frac{2M}{d^{3}}$$
$$x^{3}= \frac{d^{3}}{2}$$
$$x = \frac{d}{2^{1/3}} = \frac{14}{2^{1/3}}$$
$$x = \frac{14}{1.25}$$
$$x= 11.1 cm$$