Question

The magnetic field at the centre of a current carrying circular coil of area '$A$' is '$B$'. The magnetic moment of the coil is ($\mu_0 = $ permeability of free space)

• A. $\frac{2BA^{3/2}}{\mu_0\pi^{1/2}}$
• B. $\frac{BA^{3/2}}{\mu_0\pi}$
• C. $\frac{\mu_0\pi^{1/2} BA^2}{2}$
• D. $\frac{2BA^2}{\mu_0\pi}$

Hint:

To easily check the dimensional consistency of these types of options, remember that $M$ requires an extra $A$ ($m^2$) on top of whatever dimensions create $I$ from $B$. The $A^{3/2}$ perfectly balances the $A^{1/2}$ in the denominator from $R$.

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
We are given the magnetic field at the center of a circular coil and its area. We need to express its magnetic moment entirely in terms of the magnetic field ($B$), area ($A$), and fundamental constants.

Step 2: Key Formula or Approach:
The magnetic field at the center of a circular loop is $B = \frac{\mu_0 I}{2R}$.
The area of the loop is $A = \pi R^2$.
The magnetic moment is $M = I \times A$.

Step 3: Detailed Explanation:
First, express the radius $R$ in terms of area $A$:
$$A = \pi R^2 \implies R = \sqrt{\frac{A}{\pi}} = \frac{A^{1/2}}{\pi^{1/2}}$$
Substitute this radius into the magnetic field formula to isolate current $I$:
$$B = \frac{\mu_0 I}{2 \left( \frac{A^{1/2}}{\pi^{1/2}} \right)} = \frac{\mu_0 I \pi^{1/2}}{2A^{1/2}}$$
Solving for $I$:
$$I = \frac{2B A^{1/2}}{\mu_0 \pi^{1/2}}$$
Now, calculate the magnetic moment $M$ by multiplying the current by the area:
$$M = I \times A = \left( \frac{2B A^{1/2}}{\mu_0 \pi^{1/2}} \right) \times A$$
Combine the area terms ($A^{1/2} \times A^1 = A^{3/2}$):
$$M = \frac{2BA^{3/2}}{\mu_0\pi^{1/2}}$$

Step 4: Final Answer:
The magnetic moment is $\frac{2BA^{3/2}}{\mu_0\pi^{1/2}}$, matching option (A).