Ratio of magnetic field at the centre of a current carrying coil of radius R and at a distance of `3R` on its axis is

Options

A. `10sqrt(10)`
B. `20sqrt(10)`
C. `2sqrt(10)`
D. `sqrt(10)`

Answer

To find the ratio of the magnetic field at the center of a current-carrying coil of radius \( R \) and at a distance of \( 3R \) on its axis, we will use the formulas for the magnetic field at these two locations.

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

### Step 2: Write the formula for the magnetic field on the axis of the coil.
The magnetic field \( B_{\text{axis}} \) at a distance \( x \) from the center of the coil on its axis is given by:
\[ B_{\text{axis}} = \frac{\mu_0 I}{2(R^2 + x^2)} \]
In our case, \( x = 3R \), so we substitute this into the formula:
\[ B_{\text{axis}} = \frac{\mu_0 I}{2(R^2 + (3R)^2)} = \frac{\mu_0 I}{2(R^2 + 9R^2)} = \frac{\mu_0 I}{2(10R^2)} = \frac{\mu_0 I}{20R^2} \]

### Step 3: Calculate the ratio of the magnetic fields.
Now, we need to find the ratio \( \frac{B_{\text{center}}}{B_{\text{axis}}} \):
\[ \frac{B_{\text{center}}}{B_{\text{axis}}} = \frac{\frac{\mu_0 I}{2R}}{\frac{\mu_0 I}{20R^2}} \]
The \( \mu_0 I \) terms cancel out:
\[ = \frac{\frac{1}{2R}}{\frac{1}{20R^2}} = \frac{20R^2}{2R} = \frac{20R}{2} = 10 \]

### Conclusion
Thus, the ratio of the magnetic field at the center of the coil to the magnetic field at a distance of \( 3R \) on its axis is:
\[ \frac{B_{\text{center}}}{B_{\text{axis}}} = 10:1 \]