Due to the flow of current in a circular loop of radius `R`, the magnetic induction produced at the centre of the loop is `B`. The magnetic moment of the loop is (`mu_(0)`=permeability constant)

Options

A. `(BR^(2))/(2pi mu_(0))`
B. `(2piBR^(3))/(mu_(0))`
C. `(BR^(2))/( 2pi mu_(0))`
D. `(2pi BR^(2))/(mu_(0))`

Answer

To find the magnetic moment of a circular loop of radius \( R \) carrying a current \( I \), we can follow these steps:

### Step 1: Use the Biot-Savart Law
The magnetic induction (magnetic field) \( B \) at the center of a circular loop of radius \( R \) carrying a current \( I \) is given by the formula:
\[ B = \frac{\mu_0 I}{2R} \]
where \( \mu_0 \) is the permeability of free space.

### Step 2: Solve for Current \( I \)
From the equation for \( B \), we can rearrange it to solve for the current \( I \):
\[ I = \frac{2BR}{\mu_0} \]

### Step 3: Calculate the Area of the Loop
The area \( A \) of a circular loop is given by:
\[ A = \pi R^2 \]

### Step 4: Calculate the Magnetic Moment \( \mu \)
The magnetic moment \( \mu \) of the loop is defined as:
\[ \mu = I \cdot A \]
Substituting the expressions for \( I \) and \( A \):
\[ \mu = \left(\frac{2BR}{\mu_0}\right) \cdot \left(\pi R^2\right) \]

### Step 5: Simplify the Expression
Now, we can simplify the expression for \( \mu \):
\[ \mu = \frac{2BR \cdot \pi R^2}{\mu_0} = \frac{2\pi BR^3}{\mu_0} \]

### Final Result
Thus, the magnetic moment of the loop is:
\[ \mu = \frac{2\pi BR^3}{\mu_0} \]