Question

A 37.9 A current flows in a long, straight wire. Find the strength of the resulting magnetic field at a distance of 45.5 cm from

Answer 1

Answer:

$1.67\cdot 10^{-5} T$

Explanation:

The magnitude of the magnetic field produced by a current-carrying wire is given by the equation:

$B=\frac{\mu_0 I}{2\pi r}$

where:

$\mu_0=4\pi \cdot 10^{-7} H/m$ is the vaacuum permeability

I is the current in the wire

r is the distance from the wire

The direction of the magnetic field lines is tangential to concentric circles around the wire.

In this problem, we have:

$I=37.9 A$ is the current in the wire

$r=45.5 cm = 0.455 m$ is the distance  from the wire

Solving for B, we find the magnitude of the magnetic field:

$B=\frac{\mu_0 I}{2\pi r}=\frac{(4\pi \cdot 10^{-7})(37.9)}{2\pi (0.455)}=1.67\cdot 10^{-5} T$