The ratio of the magnetic field at the centre of a current-carrying circular coil to its magnetic moment is x. If the current and radius each of them are made three times, the new ratio will become

Options

A. 3x
B. 9x
C. `x/9`
D. `x/27`

Answer

To solve the problem, we need to find the new ratio of the magnetic field at the center of a current-carrying circular coil to its magnetic moment when both the current and the radius are tripled.

### Step-by-Step Solution:

1. **Understand the Initial Conditions**:
- Let the initial current be \( I \) and the initial radius be \( R \).
- The initial ratio of the magnetic field \( B \) at the center of the coil to its magnetic moment \( M \) is given as \( x \).

2. **Magnetic Field at the Center of the Coil**:
- The formula for the magnetic field \( B \) at the center of a circular coil is given by:
\[ B = \frac{\mu_0 I}{2R} \]
where \( \mu_0 \) is the permeability of free space.

3. **Magnetic Moment of the Coil**:
- The magnetic moment \( M \) of the coil is given by:
\[ M = I \cdot A = I \cdot \pi R^2 \]
where \( A \) is the area of the circular coil.

4. **Calculate the Initial Ratio**:
- The initial ratio \( \frac{B}{M} \) can be calculated as:
\[ \frac{B}{M} = \frac{\frac{\mu_0 I}{2R}}{I \cdot \pi R^2} = \frac{\mu_0}{2\pi R^3} \]
- We know from the problem statement that this ratio equals \( x \):
\[ \frac{\mu_0}{2\pi R^3} = x \]

5. **New Conditions**:
- Now, if the current and radius are each made three times, then:
- New current \( I' = 3I \)
- New radius \( R' = 3R \)

6. **Calculate New Magnetic Field**:
- The new magnetic field \( B' \) at the center becomes:
\[ B' = \frac{\mu_0 I'}{2R'} = \frac{\mu_0 (3I)}{2(3R)} = \frac{\mu_0 I}{2R} \]
- Notice that \( B' = B \).

7. **Calculate New Magnetic Moment**:
- The new magnetic moment \( M' \) becomes:
\[ M' = I' \cdot \pi (R')^2 = (3I) \cdot \pi (3R)^2 = 3I \cdot \pi \cdot 9R^2 = 27 \cdot (I \cdot \pi R^2) = 27M \]

8. **Calculate the New Ratio**:
- The new ratio \( \frac{B'}{M'} \) is:
\[ \frac{B'}{M'} = \frac{B}{27M} = \frac{\frac{\mu_0 I}{2R}}{27M} \]
- Since \( M = I \cdot \pi R^2 \):
\[ \frac{B'}{M'} = \frac{\frac{\mu_0 I}{2R}}{27 \cdot (I \cdot \pi R^2)} = \frac{\mu_0}{2 \cdot 27 \pi R^3} = \frac{1}{27} \cdot \frac{\mu_0}{2 \pi R^3} \]
- From our earlier calculation, we know that \( \frac{\mu_0}{2 \pi R^3} = x \):
\[ \frac{B'}{M'} = \frac{x}{27} \]

### Final Answer:
The new ratio of the magnetic field at the center of the coil to its magnetic moment is:
\[ \frac{x}{27} \]