Two parallel wires carry equal currents in the opposite directions. point a is midway between the wires, and b is an equal distance on the other side of the wires. what is the ratio of the magnitude of the magnetic field at point a to that at point b?

Complete Question:(missing part)

The current in the left wire having magnitude is  = 210 A

The Current in the right wire having magnitude is = 10 A

a.

Ratio = B 1/B 2 = 210/10 = 21

Ratio = B 1/B 2 = 210/3 X10 = 7

Explanation:

Ratio of the magnitude of the magnetic field at point A to that at point B =?

The current in the left wire having magnitude is  = 210 A

The Current in the right wire having magnitude is = 10 A

distance between two wires = d

Given that both wires are carrying equal current but in opposite direction therefore

Magnetic field linked with the left wire at point A distance =d/2= according to biot savarts law

Magnetic field linked with the right wire at point B distance =d/2

Net magnetic field added

B net = B 1 + B 2

Ratio = B 1/B 2 = 210/10 = 21

Magnetic field linked to left wire at point B(d+ d/2)

Magnetic field linked to right wire at point B(d+ d/2)

Net magnetic field subtracted

B net = B 1 - B 2

Ratio = B 1/B 2 = 210/3 X10 = 7

the answer is a. at the equator

c) up

hope this