Question

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 its cross-sectional radius is $3\, cm$ then required length of winding wire is $(B=0.2\, T)$

• A. $1.2 \times 10^2m$
• B. $4.8 \times 10^2m$
• C. $2.4 \times 10^3m$
• D. $6 \times 10^3m$

Answer 1

Answer: (C)

$2.4 \times 10^3m$

Answer 2

$B=\frac{\mu_{0} N i}{l}$; where $N =$ total number of turns, I= 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 $]$ $\Rightarrow L =2 \pi \times 3 \times 10^{-2} \times \frac{4 \times 10^{4}}{\pi}$ $=2.4 \times 10^{3} m$