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 earths magnetic field. It is rotated about its vertical diameter through $180^{\circ}$ in $0.30\, s$. Horizontal component of earth magnetic field at the place is $3\times 10^{-5}\, T$ The magnitude of current induced in the coil is approximately

• A. $4 \times 10^{-2}\,A$
• B. $8 \times 10^{-4}\,A$
• C. $8 \times 10^{-2}\,A$
• D. $1.92 \times 10^{-3}\,A$

Answer 1

The Correct Option is B

Initial flux through the coil $\phi_{i}=BA\, cos\, \theta$ $=3 \times 10^{-5}\times\pi\times(8\times10^{-2})^{2} \times cos\, 0^{\circ}$ $=192\, \pi \times 10^{-9}\,WB$ Final flux after the rotation $\phi_{f}=3\times 10^{-5}\times\pi\times(8\times10^{-2})^{2}\times cos\, 180^{\circ}$ $=-192\, \pi \times 10^{-9}\,Wb$ $\therefore$ The magnitude of induced emf is $\varepsilon=N \frac{\left|d\phi\right|}{dt}=\frac{N \left|\phi_{f}-\phi_{i}\right|}{dt}$ $=\frac{400\times\left(384\pi\times10^{-9}\right)}{0.30}$ $=1.6\times 10^{-3}\,V$ Current, $I=\frac{\varepsilon}{R}$ $=\frac{1.6\times10^{-3} }{ 2}$ $=8 \times10^{-4}\,A$