Question

A flat circular coil of $n$ turns, area $A$ and resistance $R$ is placed in a uniform magnetic field $B$. The plane of coil is initially perpendicular to $B$. When the coil is rotated through an angle of ${180}^o$ about one of its diameter, a charge $Q_1$ flows through the coil. When the same coil after being brought to its initial position, is rotated through an angle of ${360}^o$ about the same axis a charge $Q_2$ flows through it. Then $Q_2/Q_1$

• A. $1$
• B. $2$
• C. $1/2$
• D. $0$

Answer 1

The correct option is D $0$

Net charge flowing through the coil is given by $Q=\frac{\Delta \varphi }{R}$ where $R$ is the resistance

Initially the plane of ring is perpendicular to $B$, i.e Area vector $\overrightarrow{A}$ is parallel to $\overrightarrow{B}$

where $A$ is the area of the coil.

$\therefore$ initial flux ${\varphi }_i=\overrightarrow{A}.\overrightarrow{B}=AB$

Case 1) : Coil is rotated by ${180}^o$ i.e $\overrightarrow{A}$ $\parallel$ $-\overrightarrow{B}$

$\therefore$ Final flux ${\varphi }_f=-AB$

$⟹|\Delta \varphi |=|{\varphi }_f-{\varphi }_i|=2AB$

Thus $Q_1=\frac{2AB}{R}$ .............(1)

Case 2): Coil is rotated by ${360}^o$, i.e $\overrightarrow{A}$ $\parallel$ $\overrightarrow{B}$

Thus final flux ${\varphi }_f=AB$

$\therefore \Delta \varphi ={\varphi }_f-{\varphi }_i=0$

Hence $Q_2=\frac{\Delta \varphi }{R}=0$

$⟹\frac{Q_2}{Q_1}=0$