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

the wire.

Answer 1

`1.67\cdot 10^(-5) T`

Step-by-step explanation:

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

`B=(\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=(\mu_0 I)/(2\pi r)=((4\pi \cdot 10^(-7))(37.9))/(2\pi (0.455))=1.67\cdot 10^(-5) T`