A \( 400 \) turn closely wound circular coil of radius \( 10\ \mathrm{cm} \) carries a current of \( 2.2\ \mathrm{A} \). What is the magnetic field at the center of the coil?
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \( 5.526\times10^{-3}\ \mathrm{T} \)
Each turn of a circular coil gives a field \( \mu_{0}I/2R \) at its centre, and for N closely wound turns these add up (Biot-Savart law).
\( \displaystyle B = \frac{\mu_{0}NI}{2R} \)
Converting the radius to metres:
\( R = 10\ \mathrm{cm} = 0.10\ \mathrm{m} \)
\( \displaystyle B = \frac{(4\pi\times10^{-7}\ \mathrm{T\,m\,A^{-1}}) \times 400 \times (2.2\ \mathrm{A})}{2 \times 0.10\ \mathrm{m}} \)
\( B = 4\pi\times10^{-7} \times 4400\ \mathrm{T} \approx 5.526\times10^{-3}\ \mathrm{T} \)
(with \( \pi = 3.14 \))
Note: NTA picked this question from an NCERT (2026-27) example and changed the figures. Do not ignore NCERT while practising numericals.
NCERT (2026-27):
Torque on Current Loop, Magnetic Dipole – …
