Question

A curent-carrying 2.0-cm long segment of wire is inside a long solenoid adius - 40 cm - 800 turns/m, current - 50 mA). The wire segment is oriented pependicalrly to the axis of the solenoid. If the curreat in the wire segment is 12 A, what is temagnitade of the magnetic ferce on this segment?

Answer 1

Answer:

The value is  $F_{max} = 12 \muN$

Explanation:

From the question we are told that

   The  length is  $l = 2.0 \ cm = 0.02 \ m$

     The radius of the the solenoid is $r =40 \ cm$

     The number of turns per meter is  $N = 800 \ turns / m$

     The current through the solenoid  is  $I = 50 \ mA = 50*10^{-3} \ A$

      The  current through the segment is  $I_s = 12 \ A$

   Generally the magnetic force is mathematically  represented as

    $F = B * I_s l sin(\theta)$

At maximum $\theta = 90^o$

So  

   $F_{max} = B * I_s l$

Here B is the magnetic field is mathematically represented as  

      $B = \mu_o * N * I$

Here  $\mu_o$ is the permeability of free space with value

    $\mu_o = 4\pi * 10^{-7} N/A^2$

So  

      $B = 4\pi * 10^{-7} * 800 *50*10^{-3}$

=>    $B = 5.0272 *10^{-5} \ T$

So

     $F_{max} = 5.0272 *10^{-5} * 12 * 0.02$

    $F_{max} = 1.2 *10^{-5}$

      $F_{max} = 12 \muN$