At what distance on the axis, from the centre for circular current carrying coil of radius r, the magnetic field becomes 1/8 th of the magnetic field at centre?

Options

A. `sqrt(2)r`
B. `2^(3//2)r`
C. `sqrt(3)r`
D. `3sqrt(2)r`

Answer

To solve the problem, we need to find the distance \( x \) on the axis from the center of a circular current-carrying coil of radius \( r \) where the magnetic field becomes \( \frac{1}{8} \) of the magnetic field at the center of the coil.

### Step-by-Step Solution:

1. **Magnetic Field at the Center of the Coil**:
The magnetic field \( B_0 \) at the center of a circular current-carrying 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, and \( r \) is the radius of the coil.

2. **Magnetic Field at a Distance \( x \) on the Axis**:
The magnetic field \( B \) at a distance \( x \) from the center of the coil along the axis is given by:
\[ B = \frac{\mu_0 I r^2}{2(x^2 + r^2)^{3/2}} \]

3. **Setting Up the Equation**:
We want to find the distance \( x \) where the magnetic field \( B \) is \( \frac{1}{8} B_0 \):
\[ B = \frac{1}{8} B_0 \]
Substituting the expressions for \( B \) and \( B_0 \):
\[ \frac{\mu_0 I r^2}{2(x^2 + r^2)^{3/2}} = \frac{1}{8} \left(\frac{\mu_0 I}{2r}\right) \]

4. **Simplifying the Equation**:
Cancel \( \mu_0 I \) from both sides:
\[ \frac{r^2}{2(x^2 + r^2)^{3/2}} = \frac{1}{16r} \]
Cross-multiplying gives:
\[ 16r \cdot r^2 = 2(x^2 + r^2)^{3/2} \]
Simplifying further:
\[ 16r^3 = 2(x^2 + r^2)^{3/2} \]
Dividing both sides by 2:
\[ 8r^3 = (x^2 + r^2)^{3/2} \]

5. **Taking the Cube Root**:
Taking the cube root of both sides:
\[ 2r = (x^2 + r^2)^{1/2} \]

6. **Squaring Both Sides**:
Squaring both sides gives:
\[ 4r^2 = x^2 + r^2 \]
Rearranging this:
\[ x^2 = 4r^2 - r^2 = 3r^2 \]

7. **Finding \( x \)**:
Taking the square root:
\[ x = \sqrt{3}r \]

Thus, the distance \( x \) from the center of the coil on the axis where the magnetic field becomes \( \frac{1}{8} \) of the magnetic field at the center is:
\[ \boxed{\sqrt{3}r} \]