Q. A winding wire which is used to frame a solenoid can bear a maximum $10\ A$ current. If length of solenoid is $80\ cm$ and it’s cross sectional radius is $3\ cm$ then required length of winding wire is $(B = 0.2\ T)$
A) $1.2 \times 10^{2}\ m$
B) $4.8 \times 10^{2}\ m$
C) $2.4 \times 10^{3}\ m$
D) $6 \times 10^{3}\ m$
Explanation
(c) $B = \frac{\mu_{0}NI}{l}$, where $N =$ Total number of turns, $l =$ length of the solenoid
$\Rightarrow 0.2 = \frac{4\pi \times 10^{- 7} \times N \times 10}{0.8} \Rightarrow N = \frac{4 \times 10^{4}}{\pi}$
Since $N$ turns are made from the winding wire so length of the wire $(L) = 2\pi r \times N\lbrack 2\pi r =$ length of each turns]
$\Rightarrow L = 2\pi \times 3 \times 10^{- 2} \times \frac{4 \times 10^{4}}{\pi} = 2.4 \times 10^{3}m$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.