Question

Calculate the magnitude of the magnetic field at a point 16.0 cm from a long, thin conductor carrying a current of 3.50 A.

Answer 1

Answer:

The magnitude of the magnetic field is  $B = 1.392 *10^{-6} \ T$

Explanation:

From the  question we are told that

     The distance of the position that is being considered to the conductor is

         $d = 16 \ cm = 0.16 \ m$

      The current  flowing through  conductor is  $I = 3.50 \ A$

Generally the magnitude of the magnetic  field due to a current carrying conductor  is mathematically represented as

          $B = \frac{\mu_o I}{2 \pi d}$

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

substituting values

           $B = \frac{ 4\p *10^{-7} *3.50}{2 * 3.142* 0.16}$

            $B = 1.392 *10^{-6} \ T$