A circular coil with 5 turns, each of radius 7 cm, carries a current of 1.4 A. The magnetic field at the center of the coil is
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \( 2\pi\times10^{-5}\,\mathrm{T} \)
By the Biot-Savart law, the field at the centre of a circular coil of \( N \) turns and radius \( R \) is \( B = \mu_0 N I/2R \).
\( \displaystyle R = 7\,\mathrm{cm} = 0.07\,\mathrm{m} \)
\( \displaystyle B = \frac{4\pi\times10^{-7}\times5\times1.4}{2\times0.07} = 4\pi\times10^{-7}\times50 \)
\( \displaystyle B = 2\pi\times10^{-5}\,\mathrm{T} \)
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 – …
