Q.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 $B$ and carry the same current. Find $a$ in terms of $R$.

Each coil is a magnetic dipole that oscillates in the uniform field $B$ with frequency $\nu=\dfrac{1}{2\pi}\sqrt{\dfrac{m\,B}{I}}$, where $m$ is the magnetic moment and $I$ the moment of inertia. Equal $\nu$, equal $B$ and equal current require $\dfrac{m_1}{I_1}=\dfrac{m_2}{I_2}$.

With the fixed wire length $L$: $N_1(2\pi R)=L$ and $N_2(4a)=L$.

Moments (current $I_c$): $m_1=N_1 I_c\pi R^2=\dfrac{I_c L R}{2}$, $m_2=N_2 I_c a^2=\dfrac{I_c L a}{4}$. …

Concept Understanding

A current-carrying coil in a uniform field is a magnetic dipole. Displaced by a small angle it feels a restoring torque $\tau=-mB\sin\theta\approx-mB\theta$, giving simple harmonic oscillation of frequency

where $m$ is the magnetic moment and $I$ the moment of inertia about the suspension axis. Equal $\nu$ and equal $B$ for the two coils force

Step-by-Step Solution

Turns from the fixed wire length $L$:

Magnetic moments (same current $I_c$):

$m_2=N_2 I_c a^2=\frac{L}{4a}\,I_c a^2=\frac{I_c L a}{4}.$ …

Method: Equal-Frequency Condition for Two Oscillating Current Loops

This method applies to any problem comparing two current-carrying loops of different shapes, made from the same total length of wire, that oscillate in a uniform magnetic field — and asks you to relate their dimensions given that they oscillate at the same frequency.

Steps

Step 1: Recognise the physics — a current loop in a uniform field is an oscillating magnetic dipole

A planar current loop of magnetic moment $m$, displaced by a small angle $\theta$ from alignment with a uniform field $B$, feels a restoring torque $\tau=-mB\sin\theta\approx-mB\theta$. This is the same form as a torsion pendulum, so the loop performs simple harmonic angular oscillation with frequency

where $I$ is the loop's moment of inertia about the oscillation axis. Whenever two loops oscillate at the same frequency in the same field, this immediately reduces to a single condition:

Step 2: Use the fixed wire length to fix the number of turns for each geometry

If both coils are wound from the same total wire length $L$, the number of turns each geometry can make is fixed by its perimeter: $N=\dfrac{L}{\text{perimeter of one turn}}$ (e.g. $N=\dfrac{L}{2\pi R}$ for a circle of radius $R$, $N=\dfrac{L}{4a}$ for a square of side $a$). This is what lets you compare two differently shaped coils on equal footing — same wire, same current, but a different number of turns for each shape.

Step 3: Express $m$ and $I$ for each shape from standard formulas

• Magnetic moment: $m=N I_c A$, where $A$ is the area enclosed by one turn and $I_c$ is the current (the same for both coils here). …