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

Options

A. `200 mu T`
B. `250 mu T`
C. `125 mu T`
D. `75 mu T`

Answer

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

### Step 1: Understand the given information
We know:
- The radius of the circular loop, \( r = 3 \, \text{cm} = 0.03 \, \text{m} \)
- The distance from the center to the point on the axis, \( x = 4 \, \text{cm} = 0.04 \, \text{m} \)
- The magnetic field at the point on the axis, \( B_A = 54 \, \mu T = 54 \times 10^{-6} \, T \)

### Step 2: Use the formula for the magnetic field at an axial point
The magnetic field \( B \) at a point on the axis of a circular loop is given by the formula:
\[ B_A = \frac{\mu_0 I r^2}{(r^2 + x^2)^{3/2}} \]
Where:
- \( \mu_0 \) is the permeability of free space,
- \( I \) is the current flowing through the loop.

### Step 3: Calculate the magnetic field at the center of the loop
The magnetic field at the center of the loop \( B_O \) is given by:
\[ B_O = \frac{\mu_0 I}{2r} \]

### Step 4: Relate \( B_O \) to \( B_A \)
From the expressions for \( B_A \) and \( B_O \), we can derive a relationship:
\[ \frac{B_O}{B_A} = \frac{\frac{\mu_0 I}{2r}}{\frac{\mu_0 I r^2}{(r^2 + x^2)^{3/2}}} = \frac{(r^2 + x^2)^{3/2}}{2r^3} \]

### Step 5: Substitute the values
Substituting \( r = 0.03 \, \text{m} \) and \( x = 0.04 \, \text{m} \):
1. Calculate \( r^2 + x^2 \):
\[ r^2 + x^2 = (0.03)^2 + (0.04)^2 = 0.0009 + 0.0016 = 0.0025 \]
2. Calculate \( (r^2 + x^2)^{3/2} \):
\[ (0.0025)^{3/2} = (0.0025)^{1.5} = 0.00395 \]
3. Calculate \( 2r^3 \):
\[ 2r^3 = 2(0.03)^3 = 2(0.000027) = 0.000054 \]

### Step 6: Calculate the ratio
Now, substituting these into the ratio:
\[ \frac{B_O}{B_A} = \frac{0.00395}{0.000054} \approx 73.15 \]

### Step 7: Find \( B_O \)
Now, using the value of \( B_A \):
\[ B_O = B_A \cdot \frac{(r^2 + x^2)^{3/2}}{2r^3} = 54 \times 10^{-6} \cdot \frac{0.00395}{0.000054} \]
Calculating this gives:
\[ B_O \approx 54 \times 10^{-6} \cdot 73.15 \approx 3.95 \times 10^{-3} \, T = 3950 \, \mu T \]

### Final Answer
The magnetic field at the center of the loop is approximately \( 3950 \, \mu T \).