Question

A coil of resistance $40\Omega$ is connected to a galvanometer of $160\Omega$ resistance. The coil has radius $6mm$ and turns $100$. this coil is placed between the poles of a magnet such that magnetic field is perpendicular to coil is dragged out then the charge through the galvanometer is $32\mu C$. The magnetic field is:

• A. $6.55T$
• B. $5.66T$
• C. $0.655T$
• D. $0.566T$

Answer 1

The correct option is D $0.566T$

Given,

Total charge flow, $Q=32\times {10}^{-6}C$

Total resistance, $R=160+40=200\Omega$

Number of turns, $N=100$
radius of coil, $r=6mm=6\times {10}^{-3}m$

$\epsilon =-N\frac{d\varphi }{dt}$

$R\frac{dq}{dt}=-N\frac{BdA}{dt}$

$\text{for}\text{limiting}\text{values}$

$\Rightarrow R\frac{Q_{total}}{\Delta t}=-N\frac{B\Delta A}{\Delta t}$

$\Rightarrow R{Q}_{total}=-NB(0-\pi r^2)$

$\Rightarrow B=\frac{R{Q}_{total}}{N\pi r^2}=\frac{200\times 32\times {10}^{-6}}{100\times \pi \times {0.006}^2}=0.566T$

Hence, the Magnetic Field is $0.566T$