CBSE Class 12 Physics Electromagnetic Induction Solved Examples.

CBSE Class 12 Physics Electromagnetic Induction 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.4 A gramophone disc of brass of diameter 30 cm rotates horizontally at the rate of 100/3 revolutions per minute. If the vertical component of the earth's magnetic field be 0.01 weber/meter2, then the emf induced between the centre and the rim of the disc will be-

(A) 7.065 × 10^–4 V (B) 3.9 × 10^–4 V (C) 2.32 × 10^–4 V (D) none of the above.
Sol. (B) Magnetic flux passing through the disc is φ = BA = 0.01 weber/ meter² × 3.14 × (15 × 10^–2 meter)² = 7.065 × 10^–4 weber.

The line joining the centre and the circumference of the disc cuts 7.065 x 10^-4 weber flux in one round. So, the rate of cutting flux (i.e. induced emf)
= flux x number of revolutions per second
= 7.065 × 10^–4 ×  100 /60 × 3 = 3.9 × 10^–4 volt.

Ex.5 A closed coil consists of 500 turns on a rectangular frame of area 4.0 cm2 and has a resistance of 50 ohm. The coil is kept with its plane perpendicular to a uniform magnetic field of 0.2 weber/meter². The amount of charge flowing through the coil if it is turned over (rotated through 180º) will be -
(A) 1.6 × 10^–19 C (B) 1.6 × 10^–9 C (C) 1.6 × 10^–3 C (D) 1.6 × 10^–2 C

Sol. (C)
The magnetic flux passing throug each turn of a coil of area A, perpendicular to a magnetic field B is given by

Note : Rotating the coil slow or fast has no effect on the charge flown through the coil. Charge flow depends upon the total change in magnetic flux, not on the rate of change of magnetic flux.

Ex.6 A very small circular loop of area 5 × 10^–4 m², resistance 2 ohm and negligible inductance is initially coplanar and concentric with a much larger fixed circular loop of radius 0.1 m. A constant current of 1 ampere is passed in a bigger loop and the smaller loop is rotated with angular velocity ω rad/sec about a diameter. Calculate (i) the flux linked with the smaller loop, (ii) induced emf, and ((iii) induced current in the smaller loop, as a function of time (μ0 = 4π × 10^–7 V-s/A-m).

Ex.7 A copper disc of radius 0.1 m rotates about its centre with 10 revolutions per second in 'a uniform magnetic field of 0.1 tesla. The emf induced across the radius of the disc is-
(A) π/10 V (B) 2π/10 V (C) 10π mV (D) 20π mV
Sol. (C)
The induced emf between centre and rim of the rotating disc is
E =12 BωR² =1/2 × 0.1 × 2π × 10 × (0.1)²= 10π × 10^–3 volt,

Ex.8 Two rail tracks, insulated from each other and the ground, are connected to millivoltmeter. What is the reading of the milli voltmeter when a train passes at a speed of 180 km/hr along the track, given that the horizontal component of earth's magnetic field is 0.2 × 10^–4 Wb/m² and rails are separated by 1 meter.
(A) 1 mV (B) 10 mV (C) 100 mV (D) 1 V
Sol. (A)
The induced emf
E = Blv = 0.2 × 10^–4 × 1 × 180 × 1000/3600 = 0.2 × 18/3600 = 1 × 10^–3. V = 1mV

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