Question

A circular coil of wire of radius 'r' has 'n' turns and carries a current ' I '. The magnetic induction 'B' at a point on the axis of the coil of distance √3 r from its center is

• A. μ₀nI/16 r
• B. μ₀nI/32 r
• C. μ₀nI/4r
• D. μ₀nI/8 r

Answer 1

The Correct Option is A

According to the Biot-Savart Law, the magnetic field at a point due to a current-carrying loop is given by the equation:

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

Where: 

• B is the magnetic field (induction)
• \( \mu_0 \) is the permeability of free space (constant)
• n is the number of turns in the coil
• I is the current flowing through the coil
• R is the radius of the coil
• x is the distance between the point on the axis and the center of the coil

In this case, the point on the axis is a distance of \( \sqrt{3}r \) from the center of the coil. Plugging this value into the equation, we have:

\( B = \frac{\mu_0 \cdot n \cdot I \cdot r^2}{2 \cdot (r^2 + (\sqrt{3}r)^2)^{3/2}}\)

Simplifying the denominator:

\( B = \frac{\mu_0 \cdot n \cdot I \cdot r^2}{2 \cdot (r^2 + 3r^2)^{3/2}}\)

\(B = \frac{\mu_0 \cdot n \cdot I \cdot r^2}{2 \cdot (4r^2)^{3/2}}\\\)

\(B = \frac{\mu_0 \cdot n \cdot I \cdot r^2}{2 \cdot 8r^3}\\\)

\(B = \frac{\mu_0 \cdot n \cdot I}{16r}\)

Therefore, the correct option is (1) \( \frac{\mu_0 n I}{16r} \).