Question

A 21.3 A current flows in a long, straight wire. Find the strength of the resulting magnetic field at a distance of 45.7 cm from the wire.

Answer 1

Answer:

The magnetic field strength due to current flowing in the wire is9.322 x 10⁻⁶ T.

Explanation:

Given;

electric current, I = 21.3 A

distance of the magnetic field from the wire, R = 45.7 cm = 0.457 m

The strength of the resulting magnetic field at the given distance is calculated as;

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

Where;

μ₀ is permeability of free space = 4π x 10⁻⁷ T.m/A

$B = \frac{\mu_o I}{2\pi R}\\\\B = \frac{4\pi*10^{-7} *21.3}{2\pi(0.457)}\\\\B = 9.322 *10^{-6} \ T$

Therefore, the magnetic field strength due to current flowing in the wire is 9.322 x 10⁻⁶ T.