A circular coil of radius 2.0 cm carries a current of 1.0 A. If the coil has 50 turns, the magnetic field at the centre of the coil will be
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \( 1.57 \times 10^{-3}\,\mathrm{T} \)
By the Biot-Savart law, the field at the centre of a circular coil of \( N \) turns, radius \( r \), carrying current \( I \), is
\( \displaystyle B = \frac{\mu_{0} N I}{2r} \)
\( r = 2.0\,\mathrm{cm} = 2\times10^{-2}\,\mathrm{m} \)
\( \mu_{0} = 4\pi\times10^{-7}\,\mathrm{T\,m\,A^{-1}} \)
\( \displaystyle B = \frac{4\pi\times10^{-7} \times 50 \times 1.0}{2 \times 2\times10^{-2}} = 5\pi\times10^{-4}\,\mathrm{T} \)
\( \displaystyle B = 5 \times 3.14 \times 10^{-4} = 1.57\times10^{-3}\,\mathrm{T} \)
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 – …
