Two long straight wires carry currents perpendicular to the xy plane.One carries a current of 50 A and passes through the point x = 5.0 cm on the x axis.The second wire has a current of 80 A and passes through the point y = 4.0 cm on the y axis.What is the magnitude of the resulting magnetic field at the origin?

Question

Quizplus · Accepted Answer

The magnetic field at the origin due to each wire can be calculated using the formula:
B = (μ0/4π) * (I/r)
where μ0 is the magnetic constant (4π x 10^-7 Tm/A), I is the current, and r is the distance from the wire to the point of interest (in this case, the origin).

For the wire carrying 50 A, the distance from the wire to the origin is 5 cm. Plugging in the values:
B1 = (μ0/4π) * (50 A/0.05 m) = 2.0 x 10^-5 T

For the wire carrying 80 A, the distance from the wire to the origin is 4 cm. Plugging in the values:
B2 = (μ0/4π) * (80 A/0.04 m) = 4.0 x 10^-5 T

Since the wires are perpendicular to each other, the magnetic fields due to each wire add up vectorially to give the resultant magnetic field at the origin:
B = √(B1^2 + B2^2) = √[(2.0 x 10^-5)^2 + (4.0 x 10^-5)^2] ≈ 4.5 x 10^-5 T
Thus, the best choice is C) 450 μT.

Two Long Straight Wires Carry Currents Perpendicular to the Xy μ\muμ

Two Long Straight Wires Carry Currents Perpendicular to the Xy $\mu$