Question

g A solenoid 63.5 cm long has 960 turns and a radius of 2.77 cm. If it carries a current of 2.28 A, find the magnetic field along the axis at its center.Find the magnetic field on the solenoidal axis at the end of the solenoid.

Answer 1

Answer:

The  value is  $B = 0.0043 \ T$

Explanation:

From the question we are told that

   The  length of the solenoid is  $l = 63.5 = 0.635 \ m$

    The number of turns is  $N = 960 \ turns$

    The  current is  $I = 2.28 \ A$

Generally the magnetic field is mathematically represented as

      $B = \mu _o * n * I$

Where  n is the number of turn per unit length which is mathematically evaluated as

      $n = \frac{N}{l}$

     $n = \frac{960}{0.635}$

     $n = 1512 \ turns /m$

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

So  

    $B = 4\pi * 10^{-7} * 1512 * 2.28$

    $B = 0.0043 \ T$