Question

A solenoid 1.39 m long and 3.20 cm in diameter carries a current of 17.1 A. The magnetic field inside the solenoid is 26.9 mT. Find the length of the wire forming the solenoid.

Answer 1

Explanation:

It is given that,

Length of the solenoid, l = 1.39 m

Diameter of the solenoid, d = 3.2 cm = 0.032 m

Radius of solenoid, r = 0.016 m

Magnetic field inside the solenoid, $B=26.9\ mT=26.9\times 10^{-3}\ T$

We need to find the length of the wire forming the solenoid. The magnetic field inside the solenoid is given by :

$B=\mu_o nI$

n is number of turns per unit length

$B=\dfrac{\mu_o NI}{l}$

$N=\dfrac{Bl}{\mu_o I}$

$N=\dfrac{26.9\times 10^{-3}\times 1.39}{4\pi \times 10^{-7}\times 17.1}$

N = 1740.04

The total length having N loop is,

$L=2\pi r\times N$

$L=2\pi \times 0.016\times \times 1740.04$

L = 174.92 m

So, the  length of the wire forming the solenoid is 174.92 meters. Hence, this is the required solution.