Question

A circular coil of radius $10cm$, $500turns$ 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 $0.25s$. 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,
${\varphi }_i=B_{H}Acos\theta =3.0\times 10^{-5}\times \left(\pi \times 10^{-2}\right)\times cos{0}^{\circ }$
$=3\pi \times 10^{-7}Wb$
Final magnetic flux after the rotation
${\varphi }_i=3.0\times 10^{-5}\times \left(\pi \times 10^{-2}\right)\times cos180^{\circ }$
$=3\pi \times 10^{-7}Wb$
Induced emf, $\epsilon =-N\frac{d\varphi }{dt}=-\frac{N\left({\varphi }_f-{\varphi }_i\right)}{t}$
$=-\frac{500\left(-3\pi \times 10^{-7}-3\pi \times 10^{-7}\right)}{0.25}$
$=\frac{500\times \left(6\pi \times 10^{-7}\right)}{0.25}=3.8\times 10^{-3}V$
$I=\frac{\epsilon }{R}=\frac{3.8\times 10^{-3}V}{2\Omega }=1.9\times 10^{-3}A$