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

Explanation

To solve the problem, we need to understand the relationship between the magnetic dipole moment, the magnetic field at the center of the loop, and how these quantities change when the dipole moment is doubled while keeping the current constant.

### Step-by-Step Solution:

1. **Understanding Magnetic Dipole Moment (m)**:
The magnetic 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, the area \( A \) is \( \pi r^2 \). 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 a current \( I \) is given by:
\[ B = \frac{\mu_0 I}{2r} \]
where \( \mu_0 \) is the permeability of free space.

3. **Relating the Radius to the Dipole Moment**:
From the expression for \( m \), we can solve for \( r \):
\[ r = \sqrt{\frac{m}{I \pi}} \]

4. **Substituting r into the Magnetic Field Equation**:
Substituting this expression for \( r \) into the equation for \( B \):
\[ B = \frac{\mu_0 I}{2 \sqrt{\frac{m}{I \pi}}} = \frac{\mu_0 I \sqrt{I \pi}}{2 \sqrt{m}} = \frac{\mu_0 \sqrt{I^2 \pi}}{2 \sqrt{m}} \]

5. **Finding B1 and B2**:
- Let \( B_1 \) be the magnetic field when the dipole moment is \( m \):
\[ B_1 = \frac{\mu_0 \sqrt{I^2 \pi}}{2 \sqrt{m}} \]
- When the dipole moment is doubled (i.e., \( m_2 = 2m \)), the new magnetic field \( B_2 \) becomes:
\[ B_2 = \frac{\mu_0 \sqrt{I^2 \pi}}{2 \sqrt{2m}} = \frac{\mu_0 \sqrt{I^2 \pi}}{2 \sqrt{2} \sqrt{m}} \]

6. **Finding the Ratio \( \frac{B_1}{B_2} \)**:
Now, we can find the ratio:
\[ \frac{B_1}{B_2} = \frac{\frac{\mu_0 \sqrt{I^2 \pi}}{2 \sqrt{m}}}{\frac{\mu_0 \sqrt{I^2 \pi}}{2 \sqrt{2} \sqrt{m}}} = \frac{\sqrt{2}}{1} = \sqrt{2} \]

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