Phys Q In the figure below, the current in the long, straight wire is -4.80 A an
ID: 1791485 • Letter: P
Question
Phys Q
In the figure below, the current in the long, straight wire is -4.80 A and the wire lies in the plane of the rectangular loop, which carries a current 12-10.0 A. The dimensions in the figure are c-0. 100 m, a-0.150 m, and f-0.710 m. Find the magnitude and direction of the net force exerted on the loop by the magnetic field created by the wire. magnitude direction to the left X N what is the direction of the magnetic field that the long straight wire produces in the plane of the loop? what is the direction of the force that this field produces on each side of the rectangular current loop? Need Help?Explanation / Answer
The forces on the 2 horizontal parts of the loop in your diagram would yield zero.
Find the magnetic field at the vertical parts of the loop due to I1.
By Ampere's law, B dL = (µo)(I1)
(Bl)(2c) = (µo)(I1) and (Br)(2(c + a)) = (µo)(I1) (left and right portions, respectively)
(Bl) = (µo)(I1)/(2c) and (Br) = (µo)(I1)/(2(c + a))
Find the forces on the vertical parts of the loop. Since the currents are in opposite directions, the forces will also be opposite in direction.
F = (I2)(L)(Bl) - (I2)(L)(Br)
F = (I2)(L)(µo)(I1)/(2c) - (I2)(L)(µo)(I1)/(2(c + a))
F = [(I2)(L)(µo)(I1) / (2)] [1/c - 1/(c + a)]
F = [(10.0 A)(0.710 m)(1.257e-6 N/A²)(4.80 A) / (2)] [1/(0.100 m) - 1/(0.100 + 0.150 m)]
F = 4.090e-5 N