Question

A circular coil of radius $10cm,500$ turns and resistance $2\Omega$ is placed with its plane perpendicular to the horizontal component of the earth's magnetic field. If is rotated about its vertical diameter through ${180}^{\circ }m$ 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

The correct option is A $1.9\times {10}^{-3}A$

Initial magnetic flux through the coil,
${\varphi }_1=B_{H}A\cos \theta =3.0\times {10}^{-5}\times (\pi \times {10}^{-2})\times \cos 0^{\circ }$
$=3\pi \times {10}^{-7}Wb$
Final magnetic flux after the rotation
${\varphi }_f=3.0\times {10}^{-5}\times (\pi \times {10}^{-2})\times \cos {180}^{\circ }$
$=-3\pi \times {10}^{-7}Wb$.
Induced emf, $ϵ=-N\frac{d\varphi }{dt}=-\frac{N({\varphi }_f-{\varphi }_i)}{t}$
$=-\frac{500(-3\pi \times {10}^{-7}-3\pi \times {10}^{-7})}{0.25}$
$=\frac{500\times (6\pi \times {10}^{-7})}{0.25}=3.8\times {10}^{-3}V$
$I=\frac{ϵ}{R}=\frac{3.8\times {10}^{-3}V}{2\Omega }=1.9\times {10}^{-3}A$.