Question

A wire of radius 0.8 cm carries a current of 106 A that is uniformly distributed over its cross-sectional area. Find the magnetic field B at a distance of 0.07 cm from the center of the wire.

Answer 1

Answer:

The magnetic field is $B = 2.319 *10^{-3} \ T$

Explanation:

From the question we are told that

   The radius of the wire is $r = 0.8 \ cm = 0.008 \ m$

    The current is $I = 106 \ A$

    The position considered is  d = 0.07 cm = 0.0007 m

 Generally the magnetic field is mathematically represented as

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

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

So    

        $B = \frac{ 4\pi * 10^{-7} * 106 }{2 * 3.142 * \frac{0.008^2}{0.0007} }$

=>     $B = 2.319 *10^{-3} \ T$