If B is the magnetic Induction, at the centre of a circular coil of radius 't' carrying a current is 1 T, · then its value at a distance of `sqrt3 r` on the axis from the centre of the coil is

Options

A. `1/8T`
B. `1/16T`
C. `8T`
D. `1/16T`

Answer

To solve the problem, we need to find the magnetic induction (B) at a distance of \(\sqrt{3}r\) on the axis of a circular coil of radius \(r\) that carries a current, given that the magnetic induction at the center of the coil is \(1 \, \text{T}\).

### Step-by-Step Solution:

1. **Understand the Given Information:**
- The magnetic induction at the center of the coil (when \(z = 0\)) is given as \(B = 1 \, \text{T}\).
- The formula for the magnetic induction at a point on the axis of a circular coil is:
\[ B = \frac{\mu_0 I r^2}{2 (z^2 + r^2)^{3/2}} \]
- Here, \(z\) is the distance from the center of the coil along the axis.

2. **Determine the Magnetic Induction at the Center:**
- At the center of the coil, \(z = 0\):
\[ B = \frac{\mu_0 I}{2r} \]
- We know that this value is \(1 \, \text{T}\):
\[ \frac{\mu_0 I}{2r} = 1 \quad \text{(1)} \]

3. **Calculate the Magnetic Induction at Distance \(\sqrt{3}r\):**
- Now, we need to find \(B\) at \(z = \sqrt{3}r\):
\[ B' = \frac{\mu_0 I r^2}{2 ((\sqrt{3}r)^2 + r^2)^{3/2}} \]
- Simplifying the denominator:
\[ (\sqrt{3}r)^2 + r^2 = 3r^2 + r^2 = 4r^2 \]
- Therefore, we have:
\[ B' = \frac{\mu_0 I r^2}{2 (4r^2)^{3/2}} = \frac{\mu_0 I r^2}{2 \cdot 8r^3} = \frac{\mu_0 I}{16r} \]

4. **Substituting from Equation (1):**
- From equation (1), we have \(\frac{\mu_0 I}{2r} = 1\), which implies:
\[ \mu_0 I = 2r \]
- Substitute this into the expression for \(B'\):
\[ B' = \frac{2r}{16r} = \frac{1}{8} \, \text{T} \]

5. **Final Answer:**
- Thus, the magnetic induction at a distance of \(\sqrt{3}r\) on the axis from the center of the coil is:
\[ B' = \frac{1}{8} \, \text{T} \]

### Conclusion:
The value of the magnetic induction at a distance of \(\sqrt{3}r\) on the axis from the center of the coil is \(\frac{1}{8} \, \text{T}\).