A 50 cm long closely wound solenoid has 2 layers of winding of 100 turns each. The diameter of the solenoid is 1 cm. If the current in the windings is 10.0 A, the magnitude of magnetic field B inside the solenoid near its center is
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 1.6π × 10−3 T
The field inside a long solenoid, near its centre, is uniform and depends only on the number of turns per unit length and the current. Both layers carry the same current in the same sense, so all the turns count.
\( N = 2 \times 100 = 200 \text{ turns} \)
\( L = 50\,\mathrm{cm} = 0.50\,\mathrm{m} \)
\( \displaystyle n = \frac{N}{L} = \frac{200}{0.50\,\mathrm{m}} = 400\,\mathrm{m^{-1}} \)
\( B = \mu_0 nI \)
\( \displaystyle B = 4\pi\times10^{-7}\,\mathrm{T\,m\,A^{-1}} \times 400\,\mathrm{m^{-1}} \times 10.0\,\mathrm{A} \)
\( B = 1.6\pi\times10^{-3}\,\mathrm{T} \)
The diameter (1 cm) is not needed: for a solenoid much longer than its diameter, the field near the centre does not depend on the cross-section.
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 – …
