Question

A coil of radius 10 cm and resistance $40\Omega$ has 1000 turns. It is placed with its plane vertical and its axis parallel to the magnetic meridian. The coil is connected to a galvanome ter and is rotated about the vertical diameter through an angle of ${180}^{\circ }.$ Find the charge which flows through the galvanometer if the horizontal component of the earth's magnetic field is $B_H=3.0\times {10}^{-5}T.$

Answer 1

Here r = 10 cm = 0.1 m, R = 40 $\Omega$

N =1000

$\theta ={180}^{\circ },B_H=3\times {10}^{-5}T.$

$\varphi =N\left(BA\right)=NBAcos{180}^{\circ }$

= - NBA

= $-1000\times 3\times {10}^{-5}\times \pi \times 1\times 1\times {10}^{-2}$

= $3\pi \times {10}^{-4}$ weber

e = $\frac{d\varphi }{dt}=\frac{6\pi \times {10}^{-4}V}{dt}$

and i = $\frac{6\pi \times {10}^{-4}}{40dt}=\frac{4.71\times {10}^{-5}}{dt}$

Q = $\frac{4.71\times {10}^{-5}\times dt}{dt}=4.71\times {10}^{-5}C$