The magnetic field at the centre of a circular loop of radius 12.3 cm is `6.4 xx 10^(-6)` T. What will be the magnetic moment of the loop?

Answer

To find the magnetic moment of a circular loop, we can use the relationship between the magnetic field at the center of the loop and the magnetic moment. The formula for the magnetic field \( B \) at the center of a circular loop is given by:

\[ B = \frac{\mu_0 I}{2R} \]

Where:
- \( B \) is the magnetic field at the center of the loop,
- \( \mu_0 \) is the permeability of free space (\( \mu_0 = 4\pi \times 10^{-7} \, \text{T m/A} \)),
- \( I \) is the current flowing through the loop,
- \( R \) is the radius of the loop.

The magnetic moment \( \mu \) of the loop is given by:

\[ \mu = I \cdot A \]

Where \( A \) is the area of the loop, which can be calculated as:

\[ A = \pi R^2 \]

### Step 1: Calculate the area \( A \) of the circular loop.

Given:
- Radius \( R = 12.3 \, \text{cm} = 0.123 \, \text{m} \)

\[ A = \pi R^2 = \pi (0.123)^2 \]

Calculating \( A \):

\[ A \approx 3.14 \times (0.123)^2 \approx 3.14 \times 0.015129 \approx 0.0475 \, \text{m}^2 \]

### Step 2: Rearrange the formula for the magnetic field to find the current \( I \).

From the magnetic field formula:

\[ B = \frac{\mu_0 I}{2R} \]

Rearranging gives:

\[ I = \frac{2BR}{\mu_0} \]

Substituting the values:

- \( B = 6.4 \times 10^{-6} \, \text{T} \)
- \( R = 0.123 \, \text{m} \)
- \( \mu_0 = 4\pi \times 10^{-7} \, \text{T m/A} \)

Calculating \( I \):

\[ I = \frac{2 \times (6.4 \times 10^{-6}) \times 0.123}{4\pi \times 10^{-7}} \]

Calculating the numerator:

\[ 2 \times 6.4 \times 10^{-6} \times 0.123 \approx 1.57776 \times 10^{-6} \]

Now calculating \( I \):

\[ I \approx \frac{1.57776 \times 10^{-6}}{4\pi \times 10^{-7}} \approx \frac{1.57776 \times 10^{-6}}{1.25664 \times 10^{-6}} \approx 1.256 \, \text{A} \]

### Step 3: Calculate the magnetic moment \( \mu \).

Using the formula for magnetic moment:

\[ \mu = I \cdot A \]

Substituting the values:

\[ \mu = 1.256 \times 0.0475 \approx 0.0597 \, \text{A m}^2 \]

### Final Answer:

The magnetic moment of the loop is approximately \( 0.0597 \, \text{A m}^2 \).

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