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 $:
$n_{1} = \frac{L}{2\pi R}$,
$m_{1} = n_{1}iA_{1} = \left(\frac{L}{2\pi R}\right)\left(i\right)\left(\pi R^{2}\right) = \frac{LiR}{2}$
$I_{1}$ (moment of inertia of the coil $C_{1}$ about an axis through its diameter)
$= \frac{1}{2}MR^{2}$
Also, $\omega_{1} = \sqrt{\frac{m_{1}B}{I_{1}}} \quad$ (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}iA_{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)
$= \frac{1}{12}Ma^{2}$
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 $\quad \frac{LiR / 2}{\frac{1}{2}MR^{2}} = \frac{Lia / 4}{\frac{1}{12}Ma^{2}} \quad$ or $\quad \frac{Li}{MR} = \frac{3Li}{Ma}$
Hence, $a = 3R$