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.[tex]B net = \frac{\mu_{0} }{2\pi(\frac{d}{2}) } (I_{1} +I_{2} )[/tex]
Ratio = B 1/B 2 = 210/10 = 21
[tex]b.B net = \frac{\mu_{0} }{2\pi(\frac{d}{2}) } (\frac{I_{1}}{3}-I_{1}} )[/tex]
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
[tex]B = \frac{\mu_{0} \times I_{1} }{2\pi\times\frac{d}{2} }[/tex]
Magnetic field linked with the right wire at point B distance =d/2
[tex]B = \frac{\mu_{0} \times I_{2} }{2\pi\times\frac{d}{2} }[/tex]
Net magnetic field added
B net = B 1 + B 2
[tex]B net = \frac{\mu_{0} }{2\pi(\frac{d}{2}) } (I_{1} +I_{2} )[/tex]
Ratio = B 1/B 2 = 210/10 = 21
Magnetic field linked to left wire at point B(d+ d/2)
[tex]B = \frac{\mu_{0} \times I_{1} }{2\pi\times\frac{3d}{2} }[/tex]
Magnetic field linked to right wire at point B(d+ d/2)
[tex]B = \frac{\mu_{0} \times I_{2} }{2\pi\times\frac{3d}{2} }[/tex]
Net magnetic field subtracted
B net = B 1 - B 2
[tex]B net = \frac{\mu_{0} }{2\pi(\frac{d}{2}) } (\frac{I_{1} }{3}-1 } )[/tex]
Ratio = B 1/B 2 = 210/3 X10 = 7
