Question

A closed coil consists of $500$ turns on a rectangular frame of area $4.0{\text{cm}}^2$ and has a resistance of $50\Omega$. The coil is kept with its plane perpendicular to a uniform magnetic field of $0.2{\text{wb/m}}^2$, the amount of charge flowing through the coil if it is turned over (rotated through ${180}^{\circ })$

• A. $1.6\times {10}^{-3}\text{C}$
• B. $16\times {10}^{-3}\text{C}$
• C. $0.16\times {10}^{-3}\text{C}$
• D. $160\times {10}^{-3}\text{C}$

Answer 1

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

Amount of charge flowing , $\Delta Q=\frac{\Delta \varphi }{R}$

Where , $\Delta \varphi \to$ charge in flux
$R\to$ Resistance of the coil.

But, $\varphi =BA\cos \theta$

Here $\theta$ varies, so $\Delta \varphi =BA\cos {\theta }_2-BA\cos {\theta }_1$,

From the data given in the question,

$|\Delta \varphi |=2BA$

For n turns of the coil , $|\Delta \varphi |=2nBA$

Substituting the data given in the question gives,
$|\Delta \varphi |=2\times 500\times 0.2\times 4\times {10}^{-4}$

$|\Delta \varphi |=8\times {10}^{-2}\text{wb}$.

Thus, $\Delta Q=\frac{|\Delta \varphi |}{R}=\frac{8\times {10}^{-2}}{50}=1.6\times {10}^{-3}\text{C}$

Hence , option (a) is the correct answer.