Question

A solenoid of length $2\, m$ carries a current of 20 Ampere. The diameter of the solenoidis $3\, cm$. If the magnetic field inside the solenoid is $20\, mT$, then the length of wire forming the solenoid is (Assume $\mu_{0}=4 \pi \times 10^{-7} H / m$ )

• A. 100 m
• B. 125 m
• C. 175 m
• D. 150 m

Answer 1

Answer: (D)

For a solenoid,
$I=20\, A, l=2\, m, B=20\, m T=20 \times 10^{-3} T$
Diameter, $d=3\, cm =3 \times 10^{-2} m$
$\therefore $ Radius, $r=\frac{3}{2} \times 10^{-2} m$
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
$=N \times 2 \pi r=\frac{10^{4}}{2 \pi} \times 2 \pi \times \frac{3}{2} \times 10^{-2}$
$=1.5 \times 10^{2} m =150\, m$