A circular current loop of radius $0.08, m$ has $300$ turns and carries a steady current of $2.0, A$ . Find the magnitude of the magnetic field at the centre of the loop.

easy 2 marks Class 12 · Physics · Moving Charges and Magnetism

Answer

4.71 \times 1 0^{- 3}, T

The correct answer is $4.71 \times 1 0^{- 3}, T$ — Convert to decimal: $1.5 π \times 1 0^{- 3} \approx 4.71 \times 1 0^{- 3}, T$

Step-by-step solution

Compute $μ_{0} N I = (4 π \times 1 0^{- 7}) (300) (2.0) = 2.4 π \times 1 0^{- 4}, T, m$
Divide by $2 r$ to get $B = \frac{2.4 π \times 1 0 ^{- 4}}{2 ( 0.08 )} = 1.5 π \times 1 0^{- 3}, T$
Convert to decimal: $1.5 π \times 1 0^{- 3} \approx 4.71 \times 1 0^{- 3}, T$

Common mistake: Forgetting the factor

2 r

in the denominator when using

B = m u_{0} N I /2 r

Concept tested: Magnetic field of a circular loop

Published 12 September 2026 · Last updated 16 September 2026 · How this solution was prepared and checked