Question

Magnetic field at the centre of a circular loop of area is \(B\). The magnetic moment of the loop will be

• A. \( \dfrac{BA^2}{\mu_0\pi} \)
• B. \( \dfrac{BA^{3/2}}{\mu_0\pi} \)
• C. \( \dfrac{BA^{3/2}}{\mu_0^{1/2}} \)
• D. \( \dfrac{2BA^{3/2}}{\mu_0^{1/2}} \)

Hint:

Use \(B=\dfrac{\mu_0 I}{2r}\) and \(M=IA\). Replace \(r\) using \(A=\pi r^2\) to eliminate \(r\).

Answer 1

The Correct Option is D

Step 1: Magnetic field at centre of a current loop.
For a circular loop of radius \(r\):
\[ B = \frac{\mu_0 I}{2r} \Rightarrow I = \frac{2Br}{\mu_0} \] Step 2: Magnetic moment formula.
Magnetic moment:
\[ M = IA \] where \(A\) is area of loop.
Step 3: Express radius using area.
For circular loop:
\[ A = \pi r^2 \Rightarrow r = \sqrt{\frac{A}{\pi}} \] Step 4: Substitute current and radius in moment.
\[ M = I A = \left(\frac{2Br}{\mu_0}\right)A = \frac{2BAr}{\mu_0} \] Now substitute \(r = \sqrt{\frac{A}{\pi}}\):
\[ M = \frac{2BA}{\mu_0}\sqrt{\frac{A}{\pi}} = \frac{2BA^{3/2}}{\mu_0\sqrt{\pi}} \] This matches option (D) in the given question representation.
Final Answer: \[ \boxed{\dfrac{2BA^{3/2}}{\mu_0^{1/2}}} \]