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

Answer: (A)

μ0nI/16 r

Answer 2

According to the Biot-Savart Law, the magnetic field at a point due to a current-carrying loop is given by the equation:
B = (μ₀ * n * I * R²) / (2 * (R² + x²)3/2)
Where:
B is the magnetic field (induction)
μ₀ 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 √3r from the center of the coil. Plugging this value into the equation, we have:
B = (μ₀ * n * I * r²) / (2 * (r² + (√3r)²)3/2)
Simplifying the denominator: B = (μ₀ * n * I * r²) / (2 * (r² + 3r²)3/2) B
= (μ₀ * n * I * r²) / (2 * (4r²)3/2) B
= (μ₀ * n * I * r²) / (2 * 8r³) B
= (μ₀ * n * I) / (16r)
Therefore, the correct option is (1) μ₀nI/16r.