Q. There are 50 turns of a wire in every $cm$ length of a long solenoid. If $4\ ampere$ current is flowing in the solenoid, the approximate value of magnetic field along its axis at an internal point and at one end will be respectively
A) $12.6 \times 10^{- 3}weber/m^{2},\ 6.3 \times 10^{- 3}weber/m^{2}$
B) $12.6 \times 10^{- 3}weber/m^{2},\ 25.1 \times 10^{- 3}weber/m^{2}$
C) $25.1 \times 10^{- 3}weber/m^{2},\ 6.3 \times 10^{- 3}weber/m^{2}$
D) $25.1 \times 10^{- 5}weber/m^{2},\ 6.3 \times 10^{- 5}weber/m^{2}$
Explanation
(c) The magnetic field in the solenoid along its axis
(i) At an internal point $= \mu_{o}ni$
$= 4\pi \times 10^{- 7} \times 5000 \times 4 = 25.1 \times 10^{- 3}Wb/m^{2}$
[Here $n = 50\ turns/cm = 5000\ turns/m$]
(ii) At one end
$B_{end} = \frac{1}{2}B_{in} = \frac{\mu_{0}ni}{2} = \frac{25.1 \times 10^{- 3}}{2} = 12.6 \times 10^{- 3}Wb/m^{2}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.