Question

A closed coil consists of $500$ turns on a circular frame of area $4{\text{cm}}^2$ and has a resistance of $50\Omega$. The coil is kept with plane perpendicular to a uniform magnetic field of $0.2\text{Wb/m}$. The amount of charge flowing through the coil if it is turned over (rotated through ${180}^{\circ }$) will be

• A. $1.6\times {10}^{-19}\text{C}$
• B. $1.6\times {10}^{-9}\text{C}$
• C. $1.6\times {10}^{-3}\text{C}$
• D. $1.6\times {10}^{-4}\text{C}$

Answer 1

The correct option is C $1.6\times {10}^{-3}\text{C}$


The charge flown through the loop when its flux is changed by $\Delta \varphi$ & having resistance $R$ is given by,

$q=-\left(\frac{\Delta \varphi }{R}\right)=\frac{-({\varphi }_f-{\varphi }_i)}{R}$

Where,
${\varphi }_f=NBA\cos {\theta }_2\text{and}{\varphi }_i=NBA\cos {\theta }_1$

$\Rightarrow q=\frac{{\varphi }_i-{\varphi }_f}{R}$

$=\frac{NBA\cos 0^{\circ }-NBA\cos {180}^{\circ }}{R}$

$=\frac{(500\times 0.2\times 4\times {10}^{-4})(1+1)}{50}$

$=16\times {10}^{-4}$

$\Rightarrow q=1.6\times {10}^{-3}\text{C}$

Hence, option $\left(C\right)$ is correct.