Question

A coil of $40\Omega$ resistance has $100$ turns and a radius of $6mm$ is connected to an ammeter of resistance of $160\Omega$. The coil is placed perpendicular to the magnetic field. When the coil is taken out of the field, $32\mu C$ charge flows through it. The intensity of the magnetic field will be

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

Answer 1

The correct option is D

$0.566T$

Step1: Given data.

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

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

Number of turns in coil $N=100$

Radius of the Coil $=6mm=6\times {10}^{-3}m$

Step2: Finding the magnetic field intensity.

We Know that

Induced $emf$, $e=\frac{d\varphi }{dt}$ ….$\left(i\right)$

Where, $\frac{d\varphi }{dt}$is change in magnetic flux with respect to time.

Also, $emf$, $e=I\times R$ …..$\left(ii\right)$

Where, $I$ is current flowing through coil.

And, magnetic flux $\varphi =A\times B$ ….$\left(iii\right)$

Where $A$ is area of the circular coil and $B$ is magnetic filed intensity.

From equation $\left(i\right)$, $\left(ii\right)$ and $\left(iii\right)$

$\Rightarrow$ $I\times R=\frac{d\left(A\times B\right)}{dt}$

$\Rightarrow$ $\frac{dQ}{dt}\times R=A\times \frac{d\left(B\right)}{dt}$

$\Rightarrow$ $dQ\times R=A\times dB$

$\Rightarrow$ $dB=\frac{dQ\times R}{N\left(\pi \times r^2\right)}$ $\left[\because A=N\left(\pi \times r^2\right)\right]$

$\Rightarrow$ $dB=\frac{32\times {10}^{-6}C\times 200\Omega }{100\times \mathrm{\pi }\times {\left(6\times {10}^{-3}\right)}^2{m}^2}$ $\left[\because Given:Q=dQ=32\times {10}^{-6}C\right]$

$\Rightarrow$ $dB=0.566T$

Hence, option D is correct.