Question

A coil of resistance $40\Omega$ is connected to a galvanometer of $160\Omega$ resistance. The coil has radius $6$mm and turns $100$. This coil is placed between the poles of a magnet such that magnetic field is perpendicular to coil. If 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 $0.566T$

Induced emf, e= $\frac{d\varphi }{dt}$------ (1)

where, $d\varphi$ is the change in flux

As we now, emf= I$\times$R

(1)=> IR= $\frac{d\left(BA\right)}{dt}$ [Magnetic flux $\Phi$= BA]

=> $\frac{dQ}{dt}\times R=\frac{d\left(BA\right)}{dt}$

=>dB= $\frac{dQ.R}{A}$

Given, dQ= $32\times {10}^{-6}$ C

R= $(40+160)\Omega$= 200$\Omega$

$A=n\left(\pi R^2\right)$

n= no.s of turns of coil= 100 (given)

$A=100\times 3.14\times {(6\times {10}^{-3})}^2$ $m^2$

Hence, $dB=\frac{32\times {10}^{-6}\times 200}{1003.14\times 36\times {10}^{-6}}T$

= 0.566T