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Two points, point A and point B, are situated above a current-carrying wire. Point B is located at a distance R from the wire, which is twice as far from the wire as point A. By what factor is the magnetic field at point A larger or smaller than the magnetic field at point B?

1 Answer

1 vote

Answer:

By a factor of 2

Step-by-step explanation:

The strength of the magnetic field produced by a current-carrying wire is given by


B=(\mu_0 I)/(2\pi r)

where


\mu_0 is the vacuum permeability

I is the current

r is the distance from the wire

At point A, which is located at a distance
r_A from the wire, the magnetic field strength is


B_A = (\mu_0 I)/(2 \pi r_A)

Point B is located at a distance of


r_B = 2 r_A

from the wire, so the strength of the field at point B is


B_B = (\mu_0 I)/(2 \pi r_B) = (\mu_0 I)/(2 \pi (2 r_A)) = (B_A)/(2)

Or equivalently,


B_A =2B_B

So, the field at A is twice as strong as the field at B.

User Ivan Zhakov
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