Question

A coil of radius $10cm$ 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 galvanometer and is rotated about the vertical diameter through an angle of ${180}^o$. Find the charge which flows through the galvanometer if the horizontal component of the earth's magnetic field is $B=3.0\times {10}^{-5}T$

Answer 1

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

$N=1000,\theta ={180}^o,$

$B_H=3\times {10}^{-5}T;\varphi =N(B.A)=NBA\cos {180}^o$

$=-NBA=1000\times 3\times {10}^{-5}\times \pi \times 1\times {10}^{-2}=3\pi \times {10}^{-4}$

where $d\varphi =2NBA=6\pi \times {10}^{-4}weber$

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

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