Q. The inductance of a closed-packed coil of 400 turns is $8\ mH$. A current of $5\ mA$ is passed through it. The magnetic flux through each turn of the coil is
A) $\frac{1}{4\pi}\mu_{0}Wb$
B) $\frac{1}{2\pi}\mu_{0}Wb$
C) $\frac{1}{3\pi}\mu_{0}Wb$
D) $0.4\ \mu_{0}Wb$
Explanation
(a) $N\phi = Li \Rightarrow \phi = \frac{Li}{N} = \frac{8 \times 10^{- 3} \times 5 \times 10^{- 3}}{400} = 10^{- 7} = \frac{\mu_{0}}{4\pi}Wb$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.