Q. A long solenoid is formed by winding $20\ turns/cm$. The current necessary to produce a magnetic field of $20\ millitesla$ inside the solenoid will be approximately $\left( \frac{\mu_{0}}{4\pi} = 10^{- 7}\ tesla - metre/ampere \right)$
A) 8.0 $A$
B) 4.0 $A$
C) 2.0 $A$
D) 1.0 $A$
Explanation
(a) $B = \mu_{0}ni\ \Rightarrow i = \frac{B}{\mu_{0}n} = \frac{20 \times 10^{- 3}}{4\pi \times 10^{- 7} \times 20 \times 100}$
$= 7.9amp = 8amp$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.