Three long,straight,parallel wires each carry a current of 10 A in the positive x direction.If the distance between each wire and the other two is 10 cm,what is the magnitude of the magnetic force on a 20-cm length of either of the wires?
A)57 $\mu$N
B)40 $\mu$N
C)69 $\mu$N
D)50 $\mu$N
E)20 $\mu$N

Question

Quizplus · Accepted Answer

The magnitude of the magnetic force on a length of wire, assuming uniform current distribution, is given by:
F = BIL
where B is the magnetic field, I is the current, and L is the length of wire.
The magnetic field at a distance r from a long, straight wire carrying a current I is given by:
B = μ0I/(2πr)
where μ0 is the permeability of free space.
For the three wires, the distance between each wire and the other two is 10 cm (0.1 m). Therefore, the magnetic field at the center wire due to the other two wires is:
B = μ0I/(2π(0.1)) = 2×10^-4 T
The force on a 20-cm length of wire is:
F = BIL = (2×10^-4)(10)(0.2) = 4×10^-4 N
So the total force on one wire is:
Ftotal = 2F = 8×10^-4 N
Therefore, the answer is C, 69 μN.

Three Long,straight,parallel Wires Each Carry a Current of 10 a in the Positive