Question

At a certain location, Earth's magnetic field of 44 µT is horizontal and directed due north. Suppose the net field is zero exactly 11 cm above a long, straight, horizontal wire that carries a constant current. What is (a) the magnitude of the current (in A) and (b) the angle between the current direction and due north?

Answer 1

Answer:

a) The magnitude of the current is 24.19 A

b) The angle is 90°

Explanation:

a) Given:

B = magnetic field = 44 µT = 44x10⁻⁶T

r = distance = 11 cm = 0.11 m

The current is:

$B=\frac{\mu I}{2\pi r} \\I=\frac{2B\pi r}{\mu } =\frac{2*44x10^{-6}*\pi *0.11}{4\pi x10^{-7} } =24.19A$

b) Because the current flows from the west to the east, therefore, the angle is 90 degrees from the north.