Question

A flat coil, $C$ of $n$ turn, area $A$ and resistance $R$, is placed in a uniform magnetic field of magnitude $B$. The plane of the coil is initially perpendicular to $B$. The coil is rotated by an angle $\theta$ about a diameter and charge of amount $Q$ flows through it. Choose the correct alternatives.

• A. $\theta ={90}^o,Q=\left(BAn/R\right)$
• B. $\theta ={180}^o,Q=\left(2BAn/R\right)$
• C. $\theta ={180}^o,Q=0$
• D. $\theta ={360}^o,Q=0$

Answer 1

Answer: (A), (B), (D)

$\theta ={360}^o,Q=0$
$\theta ={180}^o,Q=\left(2BAn/R\right)$
$\theta ={90}^o,Q=\left(BAn/R\right)$

Current through a coil rotating with angular velocity $\omega$ is given by:

$i=\frac{nBA\omega sin\omega t}{R}$

Charge flown on rotating the coil through an angle $\theta$ is given by:

$Q=\int idt=\int \frac{i}{\omega }d\left(\omega t\right)=\int \frac{i}{\omega }d\theta$

$Q=\frac{-nBAcos\theta }{R}+\frac{nBA}{R}$

For $\theta ={90}^0$, $Q=\frac{-nBAcos{90}^0}{R}+\frac{nBA}{R}=0+\frac{nBA}{R}=\frac{nBA}{R}$

For $\theta ={180}^0$, $Q=\frac{-nBAcos{180}^0}{R}+\frac{nBA}{R}=\frac{nBA}{R}+\frac{nBA}{R}=\frac{2nBA}{R}$

For $\theta ={360}^0$, $Q=\frac{-nBAcos{360}^0}{R}+\frac{nBA}{R}=\frac{-nBA}{R}+\frac{nBA}{R}=0$