Answer:
By a factor of 2
Explanation:
The strength of the magnetic field produced by a current-carrying wire is given by
[tex]B=\frac{\mu_0 I}{2\pi r}[/tex]
where
[tex]\mu_0[/tex] is the vacuum permeability
I is the current
r is the distance from the wire
At point A, which is located at a distance [tex]r_A[/tex] from the wire, the magnetic field strength is
[tex]B_A = \frac{\mu_0 I}{2 \pi r_A}[/tex]
Point B is located at a distance of
[tex]r_B = 2 r_A[/tex]
from the wire, so the strength of the field at point B is
[tex]B_B = \frac{\mu_0 I}{2 \pi r_B} = \frac{\mu_0 I}{2 \pi (2 r_A)} = \frac{B_A}{2}[/tex]
Or equivalently,
[tex]B_A =2B_B[/tex]
So, the field at A is twice as strong as the field at B.
