Q150 / 160 · TS EAMCET 2020 (14 Sep Shift 2) · Physics

A solenoid of length $2 \mathrm{~m}$ carries a current of $20 \mathrm{~A}$. The diameter of the solenoid is $3 \mathrm{~cm}$. If the magnetic field inside the solenoid is $20 \mathrm{mT}$, then the length of wire forming the solenoid is (assume, $\mu_0=4 \pi \times 10^{-7} \mathrm{H} / \mathrm{m}$ )

  • A
    100 m
  • B
    125 m
  • C
    175 m
  • Correct answer
    D
    150 m

Solution

For a solenoid, $I=20 \mathrm{~A}, l=2 \mathrm{~m}, B=20 \mathrm{mT}=20 \times 10^{-3} \mathrm{~T}$ Diameter, $d=3 \mathrm{~cm}=3 \times 10^{-2} \mathrm{~m}$ $\therefore$ Radius, $r=\frac{3}{2} \times 10^{-2} \mathrm{~m}$ If $n$ be the number of turns per unit length in the solenoid, then $B=\mu_0 n I \Rightarrow n=\frac{B}{\mu_0 I}=\frac{20 \times 10^{-3}}{4 \pi \times 10^{-7} \times 20}=\frac{10^4}{4 \pi}$ $\therefore$ Total number of turns in the solenoid, $N=n l=\frac{10^4}{4 \pi} \times 2 \Rightarrow N=\frac{10^4}{2 \pi}$ Length of wire $=N \times$ length of wire in one turn $\begin{aligned} & =N \times 2 \pi r=\frac{10^4}{2 \pi} \times 2 \pi \times \frac{3}{2} \times 10^{-2} \\ & =1.5 \times 10^2 \mathrm{~m}=150 \mathrm{~m} \end{aligned}$

Hence, the correct answer is option: 'D'.

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