Question

There are two current carrying planar coils each made from identical wires of length L. $C_1$ is circular (radius R) and $C_2$ is square (side a). They are so constructed that they have same frequency of oscillation when they are placed in the same uniform magnetic field $\overline{B}$ and carry the same current. Find a in terms of R:

• A. $a=R$
• B. $a=2R$
• C. $a=3R$
• D. $2a=5R$

Answer 1

The correct option is C $a=3R$

Given, same frequency of small oscillation.

Torque$=\overrightarrow{\mu }\times \overrightarrow{B}$

$\Rightarrow \overrightarrow{\mu }\times \overrightarrow{B}=I\alpha$

$I\alpha =\mu B\sin \theta$

For small angle,

$I\frac{d^2\theta }{d{t}^2}=\mu B\theta$

$\Rightarrow \frac{d^2\theta }{d{t}^2}=\frac{\mu B\theta }{I}$ , i.e. compared to the Wave Equation, $\frac{d^2\theta }{d{t}^2}={\omega }^{2}theta$, $\omega =\sqrt{\frac{\mu B}{I}}$

${\omega }_1={\omega }_2$

$\Rightarrow \frac{{\mu }_1{B}_1}{I_1}=\frac{{\mu }_2{B}_2}{I_2}$

Same for $B$.

$\frac{{\mu }_1}{I_1}=\frac{{\mu }_2}{I_2}$

Given , L is turned into circles of radius r.

$\therefore n_1=\frac{L}{2\pi r},Current=I,Area=\pi r^2$

$\therefore {\mu }_1=n_{1}I\pi r^2=\frac{L}{2\pi r}\times I\times \pi r^2=\frac{LIr}{2}$

Given, L is turned into squares of side a.

$n_2=\frac{L}{4a},Current=I,Area=a^2$

${\mu }_2=n_{2}I{a}^2=\frac{L}{4a}\times I\times a^2=\frac{LIa}{4}$

$I_1=\frac{M{R}^2}{2}$, $I_2=\frac{M{a}^2}{12}$

$\Rightarrow \frac{\frac{LIr}{2}}{\frac{M{R}^2}{2}}=\frac{\frac{LIa}{4}}{\frac{M{a}^2}{12}}$

$\therefore R=\frac{a}{3}$

$\Rightarrow a=3R$