A circular coil of radius 2R is carrying current 'i' . The ratio of magnetic fields at the centre of the coil and at a point at a distance 6R from the centre of the coil on the axis of the coil is

Options

A. 10
B. `10sqrt10`
C. `20sqrt5`
D. `20sqrt10`

Answer

To solve the problem, we need to find the ratio of the magnetic fields at the center of a circular coil and at a point on the axis of the coil. Let's break it down step by step.

### Step 1: Magnetic Field at the Center of the Coil (B1)
The formula for the magnetic field at the center of a circular coil carrying current \( i \) is given by:

\[ B_1 = \frac{\mu_0 i}{2R} \]

However, in this case, the radius of the coil is \( 2R \). Therefore, we substitute \( R \) with \( 2R \):

\[ B_1 = \frac{\mu_0 i}{2 \times 2R} = \frac{\mu_0 i}{4R} \]

### Step 2: Magnetic Field at a Point on the Axis of the Coil (B2)
The formula for the magnetic field at a distance \( x \) from the center of a circular coil on its axis is given by:

\[ B = \frac{\mu_0 i R^2}{2(R^2 + x^2)^{3/2}} \]

In our case, \( R = 2R \) and \( x = 6R \). Substituting these values into the formula:

\[ B_2 = \frac{\mu_0 i (2R)^2}{2((2R)^2 + (6R)^2)^{3/2}} \]

Calculating \( (2R)^2 + (6R)^2 \):

\[ (2R)^2 = 4R^2 \quad \text{and} \quad (6R)^2 = 36R^2 \]

Thus,

\[ (2R)^2 + (6R)^2 = 4R^2 + 36R^2 = 40R^2 \]

Now substituting back into the formula for \( B_2 \):

\[ B_2 = \frac{\mu_0 i (2R)^2}{2(40R^2)^{3/2}} = \frac{\mu_0 i \cdot 4R^2}{2 \cdot (40R^2)^{3/2}} \]

Calculating \( (40R^2)^{3/2} \):

\[ (40R^2)^{3/2} = 40^{3/2} \cdot (R^2)^{3/2} = 40\sqrt{40}R^3 \]

Now substituting this back into \( B_2 \):

\[ B_2 = \frac{\mu_0 i \cdot 4R^2}{2 \cdot 40\sqrt{40}R^3} = \frac{\mu_0 i \cdot 4}{80\sqrt{40}R} = \frac{\mu_0 i}{20\sqrt{10}R} \]

### Step 3: Finding the Ratio \( \frac{B_1}{B_2} \)
Now we can find the ratio of \( B_1 \) to \( B_2 \):

\[ \frac{B_1}{B_2} = \frac{\frac{\mu_0 i}{4R}}{\frac{\mu_0 i}{20\sqrt{10}R}} = \frac{20\sqrt{10}}{4} = 5\sqrt{10} \]

### Conclusion
The ratio of the magnetic fields at the center of the coil and at a point at a distance \( 6R \) from the center of the coil on the axis of the coil is:

\[ \frac{B_1}{B_2} = 5\sqrt{10} \]