Question

A solenoid that is 133 cm long has a radius of 2.99 cm and a winding of 1740 turns; it carries a current of 3.91 A. Calculate the magnitude of the magnetic field inside the solenoid.

Answer 1

Answer:

The value is  $B = 0.0643 \ T$  

Explanation:

From the question we are told that

   The length of the solenoid is  $L = 133 \ cm = 1.33 \ m$

     The radius is  $r = 2.99 \ cm = 0.0299 \ m$

     The number of turns is $N = 1740 \ turns$

     The current it carries is  $I = 3.91 \ A$

Generally the magnitude of the magnetic field is mathematically represented as

           $B = \frac{\mu_o * N * I}{L}$

Here  $\mu_o$  is the permeability of free space with value  $\mu_o = 4\pi *10^{-7} \ N/A^2$

=>      $B = \frac{ 4\pi * 10^{-7} * 1740 * 3.91}{0.133}$  

=>      $B = 0.0643 \ T$