A 0.5 m long solenoid has 500 turns and has a flux density of 2.52 x 10−3 T at its center. The current in the solenoid is (Given, \( \mu_0 = 4\pi \times 10^{-7}\,H/m \) )
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 2.0 A
The field inside a long solenoid is \( B = \mu_0 nI \), where n is the number of turns per unit length.
\( \displaystyle n = \frac{N}{l} = \frac{500}{0.5\,\mathrm{m}} = 1000\,\mathrm{m^{-1}} \)
\( \displaystyle I = \frac{B}{\mu_0 n} \)
\( \displaystyle I = \frac{2.52\times10^{-3}\,\mathrm{T}}{4\pi\times10^{-7}\,\mathrm{H\,m^{-1}} \times 1000\,\mathrm{m^{-1}}} \)
\( \displaystyle I = \frac{2.52\times10^{-3}}{1.257\times10^{-3}}\,\mathrm{A} \approx 2.0\,\mathrm{A} \)
Note: A similar question was also asked in CUET (UG) 2026 on 30-05-2026 (Slot 2), and in CUET (UG) 2025 on 14-05-2025 (Slot 1).