Two long parallel wires each carry a current of 5.0 A directed to the east.The two wires are separated by 8.0 cm.What is the magnitude of the magnetic field at a point that is 5.0 cm from each of the wires?
A)72 $\mu$T
B)48 $\mu$T
C)24 $\mu$T
D)96$\mu$T
E)32 $\mu$T

Question

Quizplus · Accepted Answer

Using the formula for the magnetic field produced by a current-carrying wire:
B = (μ₀/4π) * (I/d)
where B is the magnetic field, I is the current, d is the distance between the wire and the point where the magnetic field is being calculated, and μ₀ is the permeability of free space (4π x 10^-7 T·m/A).

For each wire, the magnetic field at the point that is 5.0 cm away is:
B = (μ₀/4π) * (5.0 A / 0.05 m) = 2π x 10^-6 T

Since the two wires are parallel and the currents flow in the same direction, the magnetic fields add up to give the total magnetic field at the point:
Btot = 2B = 4π x 10^-6 T = 24 μT = 24 x 10^-6 T = 24 μTesla (microtesla)

Therefore, the best choice is C.

Two Long Parallel Wires Each Carry a Current of 5 μ\muμ

Two Long Parallel Wires Each Carry a Current of 5 $\mu$