MCQ
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
    $4.8 \times {10^2}\,m$
  • $2.4 \times {10^3}\,m$
  • D
    $6 \times {10^3}\,m$

Answer

Correct option: C.
$2.4 \times {10^3}\,m$
c
(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$ [ $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$

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