Question

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?

Answer 1

Answer:

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

Answer:

a.$B net = \frac{\mu_{0} }{2\pi(\frac{d}{2}) } (I_{1} +I_{2} )$

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

$b.B net = \frac{\mu_{0} }{2\pi(\frac{d}{2}) } (\frac{I_{1}}{3}-I_{1}} )$

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

$B = \frac{\mu_{0} \times I_{1} }{2\pi\times\frac{d}{2} }$

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

$B = \frac{\mu_{0} \times I_{2} }{2\pi\times\frac{d}{2} }$

Net magnetic field added

B net = B 1 + B 2

$B net = \frac{\mu_{0} }{2\pi(\frac{d}{2}) } (I_{1} +I_{2} )$

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

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

$B = \frac{\mu_{0} \times I_{1} }{2\pi\times\frac{3d}{2} }$

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

$B = \frac{\mu_{0} \times I_{2} }{2\pi\times\frac{3d}{2} }$

Net magnetic field subtracted

B net = B 1 - B 2

$B net = \frac{\mu_{0} }{2\pi(\frac{d}{2}) } (\frac{I_{1} }{3}-1 } )$

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