A long solenoid is formed by winding \( 30 \) turns per centimetre. The current required to produce a magnetic field of \( 30 \) mT inside the solenoid is
(Take \( \mu_{0} = 4\pi\times10^{-7} \) T m \( \mathrm{A^{-1}} \))
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \( \sim 8.0 \) A
Inside a long solenoid the field is uniform: \( B = \mu_{0}nI \), where n is the number of turns per unit length.
\( n = 30\ \text{turns per cm} = 30\times100 = 3000\,\mathrm{m^{-1}} \)
\( B = 30\,\mathrm{mT} = 3\times10^{-2}\,\mathrm{T} \)
\( \displaystyle I = \frac{B}{\mu_{0}n} = \frac{3\times10^{-2}}{(4\pi\times10^{-7})(3000)} \)
\( \displaystyle I = \frac{3\times10^{-2}}{3.77\times10^{-3}} \approx 7.96\,\mathrm{A} \approx 8.0\,\mathrm{A} \)