Q. A solenoid 1.5 m long and 0.4 cm in diameter possesses 10 turns/cm length. A current of 5 A falls through it. The magnetic field at the axis inside the solenoid is
A) $2\pi \times 10^{- 3}\ T$
B) $2\pi \times 10^{- 5}\ T$
C) $4\pi \times 10^{- 2}\ T$
D) $4\pi \times 10^{- 3}\ T$
Explanation
(a) Magnetic field at the axis inside the solenoid
$B = \mu_{0}ni$
Here, $n = 10\ turns\ {cm}^{- 1} = 1000\ turns\ m^{- 1},\ i = 5\ A$
$\therefore B = 4\pi \times 10^{- 7} \times 1000 \times 5$
$= 2\pi \times 10^{- 3}\ T$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.