Question

A circular coil of radius R carries current such that magnetic field at its centre is 16\(\mu\)T. Find the magnetic field on the axis at a distance of \(\sqrt{3}R\) from the centre of coil.

• A. 2\(\mu\)T
• B. 4\(\mu\)T
• C. 3\(\mu\)T
• D. 5\(\mu\)T
• E. None of these

Hint:

For distance \(x = R\sqrt{3}\), the denominator factor is always 8. For \(x = R\), it's \(2\sqrt{2}\).

Answer 1

The Correct Option is A

To solve this problem, we will first need to understand the magnetic field produced by a circular coil at its center and on its axis.

Step 1: Magnetic Field at the Center of the Coil

The magnetic field at the center of a current-carrying circular coil can be calculated using the formula:

\(B = \frac{\mu_0 I}{2R}\)

Where:

• \(B\) is the magnetic field.
• \(\mu_0\) is the permeability of free space.
• \(I\) is the current through the coil.
• \(R\) is the radius of the coil.

Given that the magnetic field at the center of the coil is 16 \(\mu \text{T}\), we have:

\(16 \times 10^{-6} = \frac{\mu_0 I}{2R}\)...(1)

Step 2: Magnetic Field on the Axis of the Coil

The magnetic field at a point on the axis at a distance \(x\) from the center of a circular coil is given by:

\(B' = \frac{\mu_0 I R^2}{2(R^2 + x^2)^{3/2}}\)

Here, we need to find the magnetic field on the axis at a distance \(\sqrt{3}R\):

\(B' = \frac{\mu_0 I R^2}{2(R^2 + (\sqrt{3}R)^2)^{3/2}}\)

Simplifying further:

\(B' = \frac{\mu_0 I R^2}{2(4R^2)^{3/2}}\)

\(B' = \frac{\mu_0 I R^2}{2(8R^3)}\)

\(B' = \frac{\mu_0 I}{16R}\)...(2)

Step 3: Relate and Compare Magnetic Fields

From equations (1) and (2), we relate B and B' using the current, which cancels out:

\(\frac{\mu_0 I}{16R} = \frac{16 \times 10^{-6} \cdot R}{\mu_0 I}\)

Therefore:

\(B' = \frac{1}{8} \times 16 \times 10^{-6} = 2 \times 10^{-6} \text{T}\)

Thus, the magnetic field at the given point on the axis is 2 \(\mu \text{T}\).

Conclusion: The correct answer is \(2\mu \text{T}\).