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 step by step, we will first establish the formulas for the magnetic field at the center of a circular coil and the magnetic moment, then derive the initial ratio, and finally calculate the new ratio after changing the current and radius.

### Step 1: Define the magnetic field at the center of a circular coil
The magnetic field \( B_0 \) at the center of a current-carrying circular coil is given by the formula:
\[ B_0 = \frac{\mu_0 I}{2R} \]
where:
- \( \mu_0 \) is the permeability of free space,
- \( I \) is the current,
- \( R \) is the radius of the coil.

### Step 2: Define the magnetic moment of the coil
The magnetic moment \( m \) of the coil is given by the formula:
\[ m = I \cdot A \]
where \( A \) is the area of the coil. The area \( A \) of a circular coil is:
\[ A = \pi R^2 \]
Thus, the magnetic moment can be expressed as:
\[ m = I \cdot \pi R^2 \]

### Step 3: Establish the initial ratio
The initial ratio of the magnetic field at the center of the coil to its magnetic moment is:
\[ \frac{B_0}{m} = \frac{\frac{\mu_0 I}{2R}}{I \cdot \pi R^2} \]
Simplifying this, we get:
\[ \frac{B_0}{m} = \frac{\mu_0}{2\pi R^3} \]
Let this ratio be equal to \( x \):
\[ x = \frac{\mu_0}{2\pi R^3} \]

### Step 4: Change the current and radius
Now, if the current \( I \) and the radius \( R \) are each made three times, we have:
- New current \( I' = 3I \)
- New radius \( R' = 3R \)

### Step 5: Calculate the new magnetic field and magnetic moment
The new magnetic field \( B_0' \) at the center of the coil becomes:
\[ B_0' = \frac{\mu_0 I'}{2R'} = \frac{\mu_0 (3I)}{2(3R)} = \frac{\mu_0 I}{2R} = B_0 \]
The new magnetic moment \( m' \) becomes:
\[ m' = I' \cdot \pi (R')^2 = (3I) \cdot \pi (3R)^2 = 3I \cdot \pi \cdot 9R^2 = 27I \cdot \pi R^2 = 27m \]

### Step 6: Establish the new ratio
The new ratio of the magnetic field at the center to the magnetic moment is:
\[ \frac{B_0'}{m'} = \frac{B_0}{27m} = \frac{\frac{\mu_0 I}{2R}}{27 \cdot (I \cdot \pi R^2)} = \frac{\mu_0}{2 \cdot 27 \cdot \pi R^3} = \frac{\mu_0}{54 \pi R^3} \]
This can be expressed in terms of \( x \):
\[ \frac{B_0'}{m'} = \frac{x}{27} \]

### Conclusion
Thus, the new ratio of the magnetic field at the center of the coil to its magnetic moment becomes:
\[ \frac{B_0'}{m'} = \frac{x}{27} \]

### Final Answer
The new ratio will become \( \frac{x}{27} \).

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