Question

If an electric wire is allowed to produce a magnetic field no larger than that of the Earth (0.50 x 10-4 T) at a distance of 15 cm from the wire, what is the maximum current the wire can carry? Express your answer using two significant figures.

Answer 1

Answer:

$I = 37.5\ A$

Explanation:

Given,

Magnetic field,B = 0.5 x 10⁻⁴ T

distance,r= 15 cm = 0.15 m

Current = ?

Using Ampere's law of magnetic field

$B = \dfrac{\mu_0I}{2\pi r}$

$I= \dfrac{B (2\pi r)}{\mu_0}$

$I= \dfrac{0.5\times 10^{-4}\times (2\pi \times 0.15)}{4\pi \times 10^{-7}}$

$I = 37.5\ A$

Current in the wire is equal to $I = 37.5\ A$