A long solenoid is formed by winding insulated copper wire at the rate of 20 turns per cm. The current that is necessary to produce a magnetic field of 20 mT inside the solenoid at its centre would be :
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 8.0 A
The field inside a long solenoid is \( B = \mu_0 n I \), where \( n \) is the number of turns per unit length.
\( n = 20\,\mathrm{cm^{-1}} = 2000\,\mathrm{m^{-1}} \)
\( B = 20\,\mathrm{mT} = 2\times10^{-2}\,\mathrm{T} \)
\( \displaystyle I = \frac{B}{\mu_0 n} = \frac{2\times10^{-2}}{4\pi\times10^{-7} \times 2000} \)
\( \displaystyle I = \frac{2\times10^{-2}}{2.513\times10^{-3}} \approx 7.96\,\mathrm{A} \approx 8.0\,\mathrm{A} \)
Note: A similar question was also asked in CUET (UG) 2026 on 31-05-2026 (Slot 2).