A 0.5 m long solenoid has 500 turns and has a flux density of \( 2.52\times10^{-3}\,\mathrm{T} \) at its centre. The current flowing in the solenoid would be
(given, \( \mu_0 = 4\pi\times10^{-7}\,\mathrm{H\,m^{-1}} \))
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 2.0 A
Inside a long solenoid the magnetic field is uniform and equal to \( B = \mu_0 n I \), where \( n \) is the number of turns per unit length.
\( \displaystyle n = \frac{N}{l} = \frac{500}{0.5\,\mathrm{m}} = 1000\,\mathrm{m^{-1}} \)
\( \displaystyle I = \frac{B}{\mu_0 n} = \frac{2.52\times10^{-3}}{4\pi\times10^{-7} \times 1000} = \frac{2.52\times10^{-3}}{1.257\times10^{-3}} \approx 2.0\,\mathrm{A} \)
Note: A similar question was also asked in CUET (UG) 2026 on 30-05-2026 (Slot 2) and on 24-05-2026 (Slot 2).