The magnetic field due to a current carrying loop of radius ` 3 cm` at a point on the axis at a distance of `4 cm` from the centre is ` 54 muT`. What will be its value at the centre of loop ?

Options

A. `125 muT`
B. `150 muT`
C. `250 muT`
D. `75 muT`

Explanation

To find the magnetic field at the center of a current-carrying circular loop, we can use the relationship between the magnetic field at a point on the axis of the loop and at the center of the loop.

### Step-by-Step Solution:

1. **Identify the Given Values**:
- Radius of the loop, \( a = 3 \, \text{cm} = 0.03 \, \text{m} \)
- Distance from the center of the loop along the axis, \( x = 4 \, \text{cm} = 0.04 \, \text{m} \)
- Magnetic field at distance \( x \), \( B = 54 \, \mu T = 54 \times 10^{-6} \, T \)

2. **Formula for Magnetic Field on the Axis**:
The magnetic field \( B \) at a distance \( x \) from the center of a circular loop is given by:
\[ B = \frac{\mu_0 I a^2}{2 (x^2 + a^2)^{3/2}} \]
where \( \mu_0 \) is the permeability of free space and \( I \) is the current.

3. **Formula for Magnetic Field at the Center**:
The magnetic field \( B' \) at the center of the loop is given by:
\[ B' = \frac{\mu_0 I}{2 a} \]

4. **Establish the Relationship**:
To find the relationship between \( B \) and \( B' \), we can set up the ratio:
\[ \frac{B'}{B} = \frac{\frac{\mu_0 I}{2 a}}{\frac{\mu_0 I a^2}{2 (x^2 + a^2)^{3/2}}} \]
Simplifying this gives:
\[ \frac{B'}{B} = \frac{(x^2 + a^2)^{3/2}}{a^3} \]

5. **Calculate \( x^2 + a^2 \)**:
\[ x^2 + a^2 = (0.04)^2 + (0.03)^2 = 0.0016 + 0.0009 = 0.0025 \, \text{m}^2 \]

6. **Calculate \( (x^2 + a^2)^{3/2} \)**:
\[ (x^2 + a^2)^{3/2} = (0.0025)^{3/2} = (0.0025)^{1.5} = 0.000125 \, \text{m}^3 \]

7. **Calculate \( a^3 \)**:
\[ a^3 = (0.03)^3 = 0.000027 \, \text{m}^3 \]

8. **Substituting Values**:
Now substituting back into the ratio:
\[ \frac{B'}{54 \times 10^{-6}} = \frac{0.000125}{0.000027} \]
Calculate the right side:
\[ \frac{0.000125}{0.000027} \approx 4.63 \]

9. **Finding \( B' \)**:
\[ B' = 54 \times 10^{-6} \times 4.63 \approx 250 \times 10^{-6} \, T = 250 \, \mu T \]

### Final Answer:
The magnetic field at the center of the loop is \( 250 \, \mu T \).

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