Question

A circular coil of radius $10 \,cm$, $500\, turns$ 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$^{\circ}$ in $0.25\, s$. The current induced in the coil is
(Horizontal component of the earth’s magnetic field at that place is $3.0 \times 10^{-5}$ T)

• A. $1.9 \times 10^{-3}\, A$
• B. $2.9 \times 10^{-3}\, A$
• C. $3.9 \times 10^{-3}\, A$
• D. $4.9 \times 10^{-3}\, A$

Answer 1

Answer: (A)

Initial magnetic flux through the coil,
$\phi_{i}=B_{H}Acos\theta=3.0\times10^{-5}\times\left(\pi\times10^{-2}\right)\times cos0^{\circ}$
$\quad =3\pi\times10^{-7}\,Wb$
Final magnetic flux after the rotation
$\phi_{i}=3.0\times10^{-5}\times\left(\pi\times10^{-2}\right)\times cos180^{\circ}$
$\quad =3\pi\times10^{-7}\,Wb$
Induced emf, $\varepsilon=-N \frac{d\phi}{dt}=-\frac{N\left(\phi_{f} -\phi_{i}\right)}{t}$
$=-\frac{500\left(-3\pi\times10^{-7}-3\pi\times10^{-7}\right)}{0.25}$
$=\frac{500\times\left(6\pi\times10^{-7}\right)}{0.25}=3.8\times10^{-3}\,V$
$I=\frac{\varepsilon}{R}=\frac{3.8\times10^{-3}\,V}{2\,\Omega}=1.9\times10^{-3}\,A$