The dipole moment of a circular loop carrying a current I, is m and the magnetic field at the centre of the loop is `B_(1)` . When the dipole moment is doubled by keeping the current constant, the magnetic field at the centre of the loop is ` B_(2)` . The ratio `(B_(1))/(B_(2))` is:

Options

A. `(1)/(sqrt(2))`
B. 2
C. `sqrt(3)`
D. `sqrt(2)`

Answer

To solve the problem, we need to understand the relationship between the dipole moment (m) of a circular loop and the magnetic field (B) at its center.

### Step-by-Step Solution:

1. **Understanding the Dipole Moment (m)**:
The dipole moment \( m \) of a circular loop carrying a current \( I \) is given by the formula:
\[ m = I \cdot A \]
where \( A \) is the area of the loop. For a circular loop of radius \( r \), the area \( A \) is:
\[ A = \pi r^2 \]
Therefore, we can express the dipole moment as:
\[ m = I \cdot \pi r^2 \]

2. **Magnetic Field at the Center of the Loop (B)**:
The magnetic field \( B \) at the center of a circular loop carrying current \( I \) is given by:
\[ B = \frac{\mu_0 I}{2r} \]
where \( \mu_0 \) is the permeability of free space.

3. **Finding the Initial Ratio \( \frac{B_1}{B_2} \)**:
Initially, we have:
\[ B_1 = \frac{\mu_0 I}{2r} \]
When the dipole moment is doubled, we have:
\[ 2m = 2(I \cdot \pi r^2) \]
Keeping the current \( I \) constant, the area must be doubled to achieve this:
\[ 2A = 2(\pi r^2) \Rightarrow A' = 2\pi r^2 \]
The new radius \( r' \) can be found from:
\[ A' = \pi (r')^2 \Rightarrow 2\pi r^2 = \pi (r')^2 \Rightarrow (r')^2 = 2r^2 \Rightarrow r' = r\sqrt{2} \]

4. **Calculating the New Magnetic Field \( B_2 \)**:
Now, substituting \( r' \) into the formula for the magnetic field:
\[ B_2 = \frac{\mu_0 I}{2r'} = \frac{\mu_0 I}{2(r\sqrt{2})} = \frac{\mu_0 I}{2\sqrt{2}r} \]

5. **Finding the Ratio \( \frac{B_1}{B_2} \)**:
Now we can find the ratio:
\[ \frac{B_1}{B_2} = \frac{\frac{\mu_0 I}{2r}}{\frac{\mu_0 I}{2\sqrt{2}r}} = \frac{1}{\frac{1}{\sqrt{2}}} = \sqrt{2} \]

### Final Answer:
The ratio \( \frac{B_1}{B_2} \) is:
\[ \frac{B_1}{B_2} = \sqrt{2} \]