Question

A solenoid used to produce magnetic fields for research purposes is 2.2 mm long, with an inner radius of 30 cmcm and 1200 turns of wire. When running, the solenoid produced a field of 1.4 TT in the center. Given this, how large a current does it carry?

Answer 1

Answer:

The current is  $I = 2042\ A$

Explanation:

From the question we are told that

    The length of the solenoid is  $l = 2.2 \ m$

    The  radius is  $r_i = 30 \ cm = 0.30 \ m$

    The number of turn is $N = 1200 \ turns$

    The  magnetic field is  $B = 1.4 \ T$

The  magnetic field produced  is mathematically represented as

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

making $I$ the subject

       $I = \frac{B * l}{\mu_o * N }$

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

 substituting values

        $I = \frac{1.4 * 2.2 }{4\pi *10^{-7} * 1200 }$

        $I = 2042\ A$