Question

There are two current carrying planar coils made each 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 $\underset{B}{\to }$ and carry the same current. Find a in terms of R.

Answer 1

Step: 1 Find the magnetic moment of both the coils.

Formula Used: m=niA

Given that two current carrying planar coils made each from identical wires of length L

Radius of the circular coil, $\left(C_1\right)$=R

Side of the square coil, $\left(C_2\right)$=a

Number of turns per unit length for circular coil, $n_1=\frac{L}{2\pi R}$

Number of turns per unit length for square coil, $n_2=\frac{L}{4n}$

Length of the wire is same i.e. $L=L_1=L_2$

Now, the magnetic moment of $C_1,m_1=n_{1}i{A}_1$

Magnetic moment of $C_2,m_1=n_{2}i{A}_2$

Substituting the values, $m_1=\frac{L.i\pi R^2}{2\pi R}$

$\Rightarrow m_1=\frac{LiR}{2}\&m_2=\frac{Li{a}^2}{4a}$

$\Rightarrow M_2=\frac{Lia}{4}$

Step: 2 Find the frequency of both the coils.

Formula Used: $f=2\pi \sqrt{\frac{l}{mB}}$

Moment of inertia of $c_1\Rightarrow I_1=\frac{M{R}^2}{2}....\left(i\right)$

Moment of inertia of $c_2\Rightarrow I_2=\frac{M{a}^2}{12}....\left(ii\right)$
(where M is the mass of coil)

Frequency of $C_1\Rightarrow f_1=2\pi \sqrt{\frac{I_1}{m_{1}B}}...\left(iii\right)$

Frequency of$C_2\Rightarrow f_1=2\pi \sqrt{\frac{I_2}{m_{2}B}}...\left(iv\right)$

Step:3 Find the value of a.According to problem, $f_1=f_2$

$2\pi \sqrt{\frac{I_1}{m_{1}B}}=2\pi \sqrt{\frac{I_2}{m_{2}B}}$

$\frac{I_1}{m_1}=\frac{I_2}{m_2}or\frac{m_2}{m_1}=\frac{I_2}{I_2}$

Substituting the values from equations, we get

$\frac{Lia.2}{4\times LiR}=\frac{M{a}^2.2}{12M{R}^2}$

$\frac{a}{2R}=\frac{a^2}{6R^2}$

3R=a

Thus, the value of a is 3R.

Final Answer: 3R