Find the magnetic field produced by a 5A current at the centre of a 50 turn coil of 8 cm diameter. Given, \( \left(\mu_0 = 4\pi\times10^{-7}\,\displaystyle\frac{\mathrm{weber}}{\mathrm{amp.m}}\right) \)
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \( 39.2 \times 10^{-4}\,\mathrm{weber/m^2} \)
By the Biot-Savart law, the field at the centre of a circular coil of \( N \) turns and radius \( R \) carrying current \( I \) is
\( \displaystyle B = \frac{\mu_0 N I}{2R} \)
\( N = 50 \)
\( I = 5\,\mathrm{A} \)
\( R = 8/2\,\mathrm{cm} = 4\,\mathrm{cm} = 0.04\,\mathrm{m} \)
\( \displaystyle B = \frac{4\pi\times10^{-7} \times 50 \times 5}{2 \times 0.04}\,\mathrm{Wb\,m^{-2}} \)
\( \displaystyle B = \frac{3.14\times10^{-4}}{0.08}\,\mathrm{Wb\,m^{-2}} = 3.925\times10^{-3}\,\mathrm{Wb\,m^{-2}} \)
\( B \approx 39.2\times10^{-4}\,\mathrm{weber/m^2} \)
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 – …
