Question

The ends of a coil, having $20\text{turns}$, area of cross-section $1{\text{cm}}^2$ and resistance $2\Omega$, are connected to a galvanometer of resistance $40\Omega$. The plane of coil is inclined at an angle ${30}^{\circ }$ to the direction of a magnetic field of intensity $1.5\text{T}$. The coil is quickly pulled out of the field, to a region of zero magnetic field. Calculate the total charge that passes through the galvanometer, during this interval.

• A. $35.7\mu \text{C}$
• B. $357\mu \text{C}$
• C. $3.57\mu \text{C}$
• D. $375\mu \text{C}$

Answer 1

The correct option is A $35.7\mu \text{C}$

Given,

$N=20\text{turns}$
$A=1{\text{cm}}^2={10}^{-4}{\text{m}}^2$
$R=2+40=42\Omega$
$B=1.5\text{T}$
$\theta ={90}^{\circ }-{30}^{\circ }={60}^{\circ }$


Total flux linked with the coil having turns $N$ and area $A$ is:

${\varphi }_1=N(\overrightarrow{B}.\overrightarrow{A})=NBA\cos \theta$

Initial flux, ${\varphi }_1=NBA\cos {60}^{\circ }$

${\varphi }_1=\frac{NBA}{2}$

When the coil is pulled out of the field, the flux becomes zero,

$\therefore$Final flux, ${\varphi }_2=0$

So, the change in flux is,

$\left|\begin{array}{c}\Delta \varphi \end{array}\right|=\frac{NBA}{2}$

The charge flowed through circuit is:

$q=\frac{\Delta \varphi }{R}=\frac{NBA}{2R}$

$\Rightarrow q=\frac{20\times 1.5\times {10}^{-4}}{2\times 42}$

$\therefore q=0.357\times {10}^{-4}C=35.7\mu \text{C}$

Hence, option ($A$) is the correct answer.

Why this Question? Short Trick: To solve these type of questions, we can use the following expression for charge flown, $q=\frac{\Delta \varphi }{R}$