Question

A magnetic field of 0.0037 T is measured at a distance of 2 cm from a long

straight current carrying wire. What is the magnitude of the current flow in the
wire?

Answer 1

Hi there!

We can use Ampère's Law to derive an expression for the magnetic field strength produced by an infinitely long current-carrying wire:

`\oint B \cdot dl = \mu_0 I_(encl)`

B = Magnetic Field Strength (T)
dl = Differential length along path

μ₀ = Permeability of Free Space (Tm/A)

I = Enclosed current (A)

The integral is a cross product, so the cosine of the angle between the magnetic field and the path of integration is used. However, for a straight current-carrying wire, the path is ALWAYS parallel to the magnetic field, so since cos(180) = 1, we can disregard the cross product.

Additionally, the path of integration is equivalent to:

`l = 2\pi r`

`B \cdot l = \mu_0 I_(encl)\\\\B * 2\pi r = \mu_0 I_(encl)\\\\B = (\mu_0 I_(encl))/(2\pi r )`

Rearrange to solve for enclosed current:

`I_(encl) = (2\pi rB)/(\mu_0) = (2\pi (0.02)(0.0037))/((4\pi *10^(-7))) = \boxed{370 A}`