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

Concept: The magnetic dipole moment of a current-carrying coil is one of the most important quantities in magnetism. It measures the strength of the magnetic effect produced by the current loop. For a coil consisting of $N$ turns, carrying current $I$, and enclosing area $A$ per turn, the magnetic moment is given by \[ M=NIA. \] Therefore, to calculate the magnetic moment, we must first determine the area enclosed by each turn of the loop. In this problem, two different coils are formed using wires of different lengths:
• A circular coil having $3$ turns and current $3\text{ A}$.
• A square coil having $2$ turns and current $2\text{ A}$. The magnetic moment of the circular coil is given. Using that information, we first determine the value of $L$, and then calculate the magnetic moment of the square coil.

Step 1: Analyze the circular coil and express its radius in terms of $L$. The second wire has total length \[ 2L. \] This wire is bent into a circular coil having \[ N_c=3 \] turns. Let the radius of each circular turn be $R$. The circumference of one turn is \[ 2\pi R. \] Since there are $3$ turns, the total wire length used is \[ 3(2\pi R). \] But this total length is given as $2L$. Hence, \[ 2L=3(2\pi R). \] Therefore, \[ 2L=6\pi R. \] Solving for $R$, \[ R=\frac{L}{3\pi}. \]

Step 2: Calculate the area enclosed by one circular turn. The area of a circle is \[ A_c=\pi R^2. \] Substituting \[ R=\frac{L}{3\pi}, \] we obtain \[ A_c = \pi\left(\frac{L}{3\pi}\right)^2. \] \[ A_c = \pi\cdot\frac{L^2}{9\pi^2}. \] \[ A_c = \frac{L^2}{9\pi}. \] Thus, the area enclosed by one circular turn is \[ A_c=\frac{L^2}{9\pi}. \]

Step 3: Form the expression for the magnetic moment of the circular coil. The circular coil has \[ N_c=3, \] current \[ I_c=3\text{ A}, \] and area per turn \[ A_c=\frac{L^2}{9\pi}. \] Using \[ M=NIA, \] we get \[ M_c = 3\times3\times\frac{L^2}{9\pi}. \] \[ M_c = \frac{L^2}{\pi}. \] But the magnetic moment is given as \[ M_c=\frac{4}{\pi}\text{ A}\cdot\text{m}^2. \] Therefore, \[ \frac{L^2}{\pi} = \frac{4}{\pi}. \] Multiplying both sides by $\pi$, \[ L^2=4. \] Hence, \[ L=2\text{ m}. \] For the remaining calculations, we use \[ L^2=4. \]

Step 4: Analyze the square coil and determine its side length. The first wire has total length \[ L. \] It is bent into a square coil having \[ N_s=2 \] turns. Let the side of each square be $a$. The perimeter of one square is \[ 4a. \] Since there are two turns, \[ L=2(4a). \] \[ L=8a. \] Thus, \[ a=\frac{L}{8}. \]

Step 5: Calculate the area enclosed by one square turn. The area of a square is \[ A_s=a^2. \] Substituting \[ a=\frac{L}{8}, \] we get \[ A_s = \left(\frac{L}{8}\right)^2. \] \[ A_s = \frac{L^2}{64}. \]

Step 6: Calculate the magnetic moment of the square coil. The square coil has \[ N_s=2, \] current \[ I_s=2\text{ A}, \] and area \[ A_s=\frac{L^2}{64}. \] Using \[ M=NIA, \] we obtain \[ M_s = 2\times2\times\frac{L^2}{64}. \] \[ M_s = \frac{4L^2}{64}. \] \[ M_s = \frac{L^2}{16}. \] Substituting \[ L^2=4, \] we get \[ M_s = \frac{4}{16}. \] \[ M_s = \frac{1}{4}. \] Therefore, \[ M_s=0.25\text{ A}\cdot\text{m}^2. \] Final Answer: The magnetic moment of the square loop is \[ \boxed{0.25\text{ A}\cdot\text{m}^2}. \] Hence, the correct option is \[ \boxed{\text{(D)}\;0.25\text{ A}\cdot\text{m}^2}. \]