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 $\vec{B}$ 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

Answer: (C)

For coil $C_1$:
$n_1=\frac{L}{2\pi R}$,
$m_1=n_{1}i{A}_1=\left(\frac{L}{2\pi R}\right)\left(i\right)\left(\pi R^2\right)=\frac{LiR}{2}$
$I_1$ (moment of inertia of the coil $C_1$ about an axis through its diameter)
$=\frac{1}{2}M{R}^2$
Also, ${\omega }_1=\sqrt{\frac{m_{1}B}{I_1}}$ (as $T_1=2\pi \sqrt{\frac{I_1}{mB}}$)
For the coil $C_2$ :
As $n_2=\frac{L}{4a}$,
$m_2=n_{2}i{A}_2=\left(\frac{L}{4a}\right)\left(i\right)\left(a^2\right)=\frac{Lia}{4}$
$I_2$ (moment of inertia of coil $C_2$ about an axis through its centre and parallel to one of its sides)
$=\frac{1}{12}M{a}^2$
Also, ${\omega }_2=\sqrt{\frac{m_{2}B}{I_2}}$
Since ${\omega }_1={\omega }_2$,
$\frac{m_1}{I_1}=\frac{m_2}{I_2}$
or $\frac{LiR/2}{\frac{1}{2}M{R}^2}=\frac{Lia/4}{\frac{1}{12}M{a}^2}$ or $\frac{Li}{MR}=\frac{3Li}{Ma}$
Hence, $a=3R$