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

The magnetic field is
`B = 2.319 *10^(-3) \ T`

Step-by-step 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 = (\mu_o * I)/(2\pi * (r^2)/(d) )`

Here
`\mu_o` is the permeability of free space with value
`4\pi * 10^(-7) N/A^2`

So

`B = ( 4\pi * 10^(-7) * 106 )/(2 * 3.142 * (0.008^2)/(0.0007) )`

=>
`B = 2.319 *10^(-3) \ T`