Question

A solenoid of length $2m$ carries a current of 20 Ampere. The diameter of the solenoidis $3cm$. If the magnetic field inside the solenoid is $20mT$, 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,
$I=20A,l=2m,B=20mT=20\times 10^{-3}T$
Diameter, $d=3cm=3\times 10^{-2}m$
$\therefore$ Radius, $r=\frac{3}{2}\times 10^{-2}m$
If $n$ be the number of turns per unit length in the solenoid, then
$B={\mu }_{0}nI\Rightarrow n=\frac{B}{{\mu }_{0}I}$
$=\frac{20\times 10^{-3}}{4\pi \times 10^{-7}\times 20}=\frac{10^4}{4\pi }$
$\therefore$ Total number of turns in the solenoid,
$N=nl=\frac{10^4}{4\pi }\times 2$
$\Rightarrow N=\frac{10^4}{2\pi }$
Length of wire $=N\times$ length of wire in one turn
$=N\times 2\pi r=\frac{10^4}{2\pi }\times 2\pi \times \frac{3}{2}\times 10^{-2}$
$=1.5\times 10^{2}m=150m$