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

Answer: (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, $I=\frac{\varepsilon}{R}$
$=\frac{1.6\times10^{-3} }{ 2}$
$=8 \times10^{-4}\,A$