Question

A thin conducting wire of length $L$ carrying a current of $2\text{ A}$ is bent into a square loop of $2$ turns and another thin conducting wire of length $2L$ carrying a current of $3\text{ A}$ is bent into a circular loop of $3$ turns. If the magnetic moment of the circular loop is $\frac{4}{\pi}\text{ A}\cdot\text{m}^2$, then the magnetic moment of the square loop is:

• A. $0.5\pi\text{ A}\cdot\text{m}^2$
• B. $0.5\text{ A}\cdot\text{m}^2$
• C. $0.25\pi\text{ A}\cdot\text{m}^2$
• D. $0.25\text{ A}\cdot\text{m}^2$

Hint:

For current-carrying coils, \[ M=NIA. \] Whenever the wire length is given, first express the dimensions of the figure (radius, side length, etc.) in terms of the total wire length. Then calculate the enclosed area and substitute into the magnetic moment formula. This approach avoids unnecessary calculations and is especially useful in JEE and NEET problems involving multiple-turn coils.

Answer 1

The Correct Option is D

Step 1: Recall the magnetic moment of a coil.
For $N$ turns carrying current $I$ and enclosing area $A$ per turn, $M = NIA$. We must find the area from how each wire is bent.
Step 2: Use the circular coil to find L.
Wire of length $2L$ is wound into $3$ circular turns, so $2L = 3(2\pi R)$, giving $R = \dfrac{L}{3\pi}$. Its area per turn is $A_c = \pi R^2 = \pi \dfrac{L^2}{9\pi^2} = \dfrac{L^2}{9\pi}$.
Step 3: Apply the known circular moment.
$M_c = N_c I_c A_c = 3(3)\dfrac{L^2}{9\pi} = \dfrac{L^2}{\pi}$. This equals $\dfrac{4}{\pi}$, so $L^2 = 4$ and $L = 2\,m$.
Step 4: Geometry of the square coil.
Wire of length $L$ is wound into $2$ square turns. Total perimeter $= 2(4a) = 8a = L$, so the side is $a = \dfrac{L}{8}$, and area per turn $A_s = a^2 = \dfrac{L^2}{64}$.
Step 5: Magnetic moment of the square coil.
$M_s = N_s I_s A_s = 2(2)\dfrac{L^2}{64} = \dfrac{4L^2}{64} = \dfrac{L^2}{16}$.
Step 6: Substitute and conclude.
With $L^2 = 4$, $M_s = \dfrac{4}{16} = 0.25\,A\cdot m^2$, which is option (2).
\[ \boxed{M_s = 0.5\,A\cdot m^2} \]