Question

calculate the current required to create a 0 .85-mT magnetic field at the center of a loop of wire of diameter 12 cm

Answer 1

Answer:

81.2 A

Explanation:

The magnetic field at the center of a current-carrying loop is given by

$B=\frac{\mu_0 I}{2R}$

where

$\mu_0 = 4\pi \cdot 10^{-7} Tm/A$ is the vacuum permeability

I is the current in the loop

R is the radius of the loop

In this problem, we have:

d = 12 cm is the diameter of the loop, so its radius is

R = d/2 = 12/2 = 6 cm = 0.06 m

$B=0.85 mT=0.85\cdot 10^{-3}T$ is the magnetic field at the center

Therefore, the current in the loop must be:

$I=\frac{2RB}{\mu_0}=\frac{2(0.06)(0.85\cdot 10^{-3}}{4\pi \cdot 10^{-7}}=81.2 A$