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 can follow these steps:

### Step 1: Magnetic Field at the Center of the Coil
The formula for the magnetic field \( B \) at the center of a circular coil carrying current \( I \) is given by:

\[ B_{\text{center}} = \frac{\mu_0 I}{2R} \]

where:
- \( \mu_0 \) is the permeability of free space,
- \( I \) is the current flowing through the coil,
- \( R \) is the radius of the coil.

### Step 2: Magnetic Field at Point P on the Axis
The magnetic field at a point on the axis of the coil at a distance \( x \) from the center is given by the formula:

\[ B_{P} = \frac{\mu_0 I R^2}{2(R^2 + x^2)^{3/2}} \]

In this case, \( x = 3R \). Therefore, we substitute \( x \) into the equation:

\[ B_{P} = \frac{\mu_0 I R^2}{2(R^2 + (3R)^2)^{3/2}} = \frac{\mu_0 I R^2}{2(R^2 + 9R^2)^{3/2}} = \frac{\mu_0 I R^2}{2(10R^2)^{3/2}} \]

### Step 3: Simplifying the Magnetic Field at Point P
Now we simplify \( B_{P} \):

\[ B_{P} = \frac{\mu_0 I R^2}{2(10^{3/2} R^3)} = \frac{\mu_0 I}{2 \cdot 10^{3/2} R} \]

### Step 4: Finding the Ratio of the Magnetic Fields
Now we find the ratio of the magnetic field at the center to that at point \( P \):

\[ \text{Ratio} = \frac{B_{\text{center}}}{B_{P}} = \frac{\frac{\mu_0 I}{2R}}{\frac{\mu_0 I}{2 \cdot 10^{3/2} R}} \]

This simplifies to:

\[ \text{Ratio} = \frac{1}{\frac{1}{10^{3/2}}} = 10^{3/2} = 10 \sqrt{10} \]

### Conclusion
Thus, the ratio of the magnetic field at the center of the coil to that at a distance of \( 3R \) on its axis is:

\[ \text{Ratio} = 10 \sqrt{10} \]