CBSE Class 12 Physics Magnetism Solved Examples.

CBSE Class 12 Physics Magnetism Solved Examples.. Please refer to the examination notes which you can use for preparing and revising for exams. These notes will help you to revise the concepts quickly and get good marks.

Ex.1 A north pole of strength 50 Am and south pole of strength 100 Am are separated by a distance of 10 cm in air. Find the force between them. Also find the force when the distance between them is doubled.
Sol. Force between magnetic poles in air is given

Ex.2 Calculate the strength of the magnetic field at a distance of 20 cm from a pole of strength40 Am in air. Find the induction at the same point.
Sol. Strength of a magnetic field due to a pole of strength m is given by

Ex.3 A bar magnet of length 0.2 m and pole strength 5 Am is kept in a uniform magnetic induction field of strength 15 Wbm^–2 making an angle of 30° with the field. Find the couple acting on it.
Sol. Couple acting on a bar magnet of dipole moment M when placed in a magnetic field,is given by τ = MB sin θ where θ is the angle made by the axis of magnet with the direction of field.

Ex.4 A short magnet of length 4 cm is kept at a distance of 20 cm to the east of a compass box such that is axis is perpendicular to the magnetic meridian. If the deflection produced is 45°, find the pole strength (H = 30 Am^–1)

Sol. The compass box will be on the axial line of the magnet,

Ex.5 A barmagnet of magnetic moment 1.5 JT^–1 lies aligned with the direction of a uniform magnetic field of 0.22 T.
(a) What is the amount of work required to turn the magnet so as to align is magnetic moment
(i) normal to the field direction,
(ii) opposite to the field direction ?
(i) What is the torque on the magnet in cases
(i) and (ii) ?
Sol. Given that : M 1.5 JT^–1 and B = 0.22 T
(a) The potential energy of the magnet when its magnetic moment vector →M makes an angle θ with the magnetic field is givenby, U = – MB cos θ When the magnetic moment is aligned with the field i.e., θ = 0, the potential energy is U₀ = – MB cos 0° = – MB When the magnetic moment is normal to the field direction, i.e. θ = 90°, the potential energy is U₀ = – MB cos 90° = 0.When the magnetic moment is opposite to the field direction i.e., θ = 180°, the potential energy is U₁₈₀ = – MB cos 180° = + MB
(i) Hence, work done to turn the magnet to align its magnetic dipole moment normal to the field W₀ – 90 = U₉₀ – U0 = 0 – (–MB) = MB = 1.5 × 0.22 J = 0.33 J

(ii) Similarly, work done to turn the magnet to align its magnetic dipole moment opposite to thefield direction. W0 – 180 = U₁₈₀  – U0 = MB – (–MB) = 2 MB = 2 × 0.33 = 0.66 J.
(b) Torque on the magnet in case (i) :τ1 = MB sin 90° = MB = 0.33 N-m Similarly, torque on the magnet in case
(ii) τ2 = MB sin 180° = 0 N - m.

Ex.6 A long straight horizontal cable carries a current of 2.5 A in the direction 10° south of west to 10° north of east. The magnetic meridian of the place happens to be 10° west of the geographic meridian. The earth’s magnetic field at the location is 0.33 Gauss, and the angle of dip is zero. Locate the line of neutral points. (Ignore the thickness of the cable).
Sol. The situation is shown in figure. The horizontal component of earth’s magnetic field at the location of the cable (angle of dip θ = 0 is)

Ex.7 A 10 cm long bar magnet of magnetic moment 1.34 Am² is placed in the magnetic meridian with its south pole pointing geographical south. The neutral point is obtained at a distance of 15 cm from the centre of the magnet. Calculate the horizontal component of earth’s magnetic field.
Sol. As the magnet is placed with its south pole pointing south, hence the neutral point lies on the equatorial line. At the neutral point, the magnetic field B due to the magnet becomes equal and opposite to horizontal component of earth’s magnetic field i.e., BH. Hence, if M be magnetic dipole moment of the magnet of length 2l and r the distance of the neutral point from its centre, then

Ex.8 A circular coil of 16 turns and radius 10 cm carries a current of 0.75 A and rest with its plane normal to an external magnetic field of 5.0 × 10^–2 T. The coil is free to rotate about an axis in its plane perpendicular to the magnetic field. When the coil is slightly turned and released, it oscillates about its stable equilibrium position with a frequency of 2.0 s^–1 Compute the moment of inertia of the coil about its axis of rotation.     

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