Question

A wire carries a current of 4.2 A at what distance from the wire does the magnetic field have a magnitude of 1.3×10^ -5 t

Answer 1

the distance is 6.46 cm.

Step-by-step explanation:

Given

current in the wire, I = 4.2 A

magnitude of the magnetic field, B = 1.3 x 10⁻⁵ T

The distance from the wire is determined by using Biot-Savart Law;

`B = (\mu_o I)/(2\pi r) \\\\r = (\mu_o I)/(2\pi B)`

Where;

r is the distance from the wire where the magnetic field is experienced

`r = (\mu_o I)/(2\pi B)\\\\r = (4\pi * 10^(-7) * 4.2 )/(2\pi * 1.3 * 10^(-5))\\\\r = 0.0646 \ m\\\\r = 6.46 \ cm`

Therefore, the distance is 6.46 cm.