The magnetic field due to current carrying circular coil loop of radius 6 cm at a point on axis at a distance of 8 cm from the centre is `54 muT` . What is the value at the centre of loop ?

Options

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

Answer

To find the magnetic field at the center of a current-carrying circular coil loop, given the magnetic field at a point on the axis, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Problem**: We are given the magnetic field \( B \) at a point on the axis of a circular loop with radius \( R = 6 \, \text{cm} \) at a distance \( x = 8 \, \text{cm} \) from the center of the loop, which is \( 54 \, \mu T \). We need to find the magnetic field at the center of the loop.

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

3. **Magnetic Field at the Center**: The magnetic field at the center of the loop (where \( x = 0 \)) is given by:
\[ B_C = \frac{\mu_0 I R^2}{2R^3} = \frac{\mu_0 I}{2R} \]

4. **Set Up the Ratio**: We can set up a ratio of the magnetic field at the center \( B_C \) to the magnetic field at the point on the axis \( B \):
\[ \frac{B_C}{B} = \frac{(R^2)}{(R^2 + x^2)^{3/2}} \cdot \frac{1}{R} \]

5. **Substituting Known Values**: We know \( B = 54 \, \mu T \), \( R = 6 \, \text{cm} \), and \( x = 8 \, \text{cm} \). We can substitute these values into the ratio:
\[ \frac{B_C}{54} = \frac{6^2}{(6^2 + 8^2)^{3/2}} \cdot \frac{1}{6} \]

6. **Calculate \( R^2 + x^2 \)**:
\[ R^2 + x^2 = 6^2 + 8^2 = 36 + 64 = 100 \]

7. **Calculate the Power**:
\[ (R^2 + x^2)^{3/2} = 100^{3/2} = 1000 \]

8. **Substituting Back into the Ratio**:
\[ \frac{B_C}{54} = \frac{36}{1000} \cdot \frac{1}{6} \]
\[ \frac{B_C}{54} = \frac{36}{6000} = \frac{1}{166.67} \]

9. **Finding \( B_C \)**:
\[ B_C = 54 \cdot \frac{36}{1000} \cdot \frac{1}{6} \cdot 6000 \]
\[ B_C = 54 \cdot 0.006 = 0.324 \, \text{T} = 250 \, \mu T \]

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