A solenoid of length 30 cm and 3000 turns carries a current of 10 A. The value of the magnetic field \( (\vec{B}) \) at the centre of solenoid is: (Given \( \mu_0 = 4\pi \times 10^{-7}\,\mathrm{Wb/A}\text{-}\mathrm{m} \))
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \( 0.13\,\mathrm{Wb/m^2} \)
Inside a long solenoid the field is uniform and is \( B = \mu_0 n I \), where \( n \) is the number of turns per unit length.
\( \displaystyle L = 30\,\mathrm{cm} = 0.30\,\mathrm{m} \)
\( \displaystyle n = \frac{N}{L} = \frac{3000}{0.30} = 10^{4}\,\mathrm{m^{-1}} \)
\( \displaystyle B = 4\pi\times10^{-7} \times 10^{4} \times 10 = 4\pi\times10^{-2}\,\mathrm{Wb\,m^{-2}} \)
\( \displaystyle B = 0.1257\,\mathrm{Wb\,m^{-2}} \approx 0.13\,\mathrm{Wb\,m^{-2}} \)
Note: NTA picked this question from an NCERT (2026-27) example and changed the figures. Do not ignore NCERT while practising numericals.
NCERT (2026-27):
The Solenoid – …
