The magnetic field at the centre of a closely wound circular coil of 100 turns, radius 22 cm, carrying a current of 3.5 A will be
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \( 1 \times 10^{-3}\,\mathrm{T} \)
By the Biot-Savart law, the field at the centre of a circular coil of \( N \) closely wound turns and radius \( R \) is
\( \displaystyle B = \frac{\mu_0 N I}{2R} \)
\( N = 100 \)
\( I = 3.5\,\mathrm{A} \)
\( R = 22\,\mathrm{cm} = 0.22\,\mathrm{m} \)
\( \mu_0 = 4\pi\times10^{-7}\,\mathrm{T\,m\,A^{-1}} \)
\( \displaystyle B = \frac{4\pi\times10^{-7} \times 100 \times 3.5}{2 \times 0.22} = \frac{1400\pi\times10^{-7}}{0.44} \)
Taking \( \pi = 22/7 \), \( 1400\pi = 4400 \):
\( \displaystyle B = \frac{4400\times10^{-7}}{0.44} = 1\times10^{-3}\,\mathrm{T} \)
Note: NTA picked this question from an NCERT (2026-27) exercise and changed the figures. Do not ignore NCERT while practising numericals.
NCERT (2026-27):
Exercises – …
