Q. A long solenoid of 50 cm length having 100 turns carries a current of 2.5 A. The magnetic field at the centre of the solenoid is : ($\mu_0 = 4\pi \times 10^{-7}$ T m $A^{-1}$)
A) $3.14 \times 10^{-4}$ T
B) $6.28 \times 10^{-5}$ T
C) $3.14 \times 10^{-5}$ T
D) $6.28 \times 10^{-4}$ T
Explanation
[Solution] n = 100/0.5 = 200 turns/m; $B = \mu_0 nI = 4\pi \times 10^{-7} \times 200 \times 2.5 = 6.28 \times 10^{-4}$ T. Answer (4).
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.