Question

A circular coil of radius 8 cm, 400 turns and resistance 2$\Omega$ is placed with its plane perpendicular to the horizontal component of the earth’s magnetic field. It is rotated about its vertical diameter through 180° in 0.30 s. Horizontal component of earth magnetic field at the place is 3 × 10-5 T. The magnitude of current induced in the coil is approximately

• A. 4 × 10-2 A
• B. 8 × 10-4 A
• C. 8 × 10-2 A
• D. 1.92 × 10-3 A

Answer 1

Answer: (B)

8 × 10-4 A

Answer 2

Initial flux through the coil

$\varphi_{i} = BA\cos\theta$

$= 3 \times 10^{- 5} \times \pi \times (8 \times 10^{- 2})^{2} \times \cos 0{^\circ}$

$= 192\pi \times 10^{- 9}Wb$

Final flux after the rotation

$\varphi_{f} = 3 \times 10^{- 5} \times \pi \times (8 \times 10^{- 2})^{2} \times \cos 180{^\circ}$

$= - 192\pi \times 10^{- 9}Wb$

$\therefore$The magnitude of induced emf is

$\varepsilon = N\frac{|d\varphi|}{dt} = \frac{N|\varphi_{f} - \varphi_{i}|}{dt}$

$= \frac{400 \times (384\pi \times 10^{- 9})}{0.30} = 1.6 \times 10^{- 3}V$

Current $I = \frac{\varepsilon}{R} = \frac{1.6 \times 10^{- 3}}{2} = 8 \times 10^{- 4}A$