Two insulated current-carrying wires (wire 1 and wire 2) are bound together with

Question

Two insulated current-carrying wires (wire 1 and wire 2) are bound together with wire ties to form a two-wire unit. The wires are 2.68 m long and are stretched out horizontally parallel to each other wire 1 carries a current of h·8.00 A and the other wire carnes a current 12 in the opposite direction. The two-wire unit is placed in a uniform magnetic field of magnitude 0.400 T such that the angle between the direction of 1, and the magnetic field is 70.00. While we don't know the current in wire 2, we do know that it is smaller than the current in wire 1. If the magnitude of the net force experienced by the two-wire unit is 3.50 N, determine the current in wire 4.03 How does the direction of the magnetic force exerted on wire 1 compare to the direction of the magnetic force exerted on wire 2? How is the magnitude of the net force acting on the two-wire knit related to the magnitude of the magnetic force acting on each wire? What physical quantities determine the magnitude of the magnetic force acting on each wire? A

Explanation / Answer

Given,

L = 2.68 m ; I1 = 8 A ; I2 = ? ; theta = 70 deg ; B = 0.4 T ; F = 3.5 N

Since the wires carry current in apposite direction the force between them will be repulsive.

Fnet = B I L sin(theta)

I = (8 - I2)

3.5 = 0.4 x (8 - I2) 2.68 x sin70

8 - I2 = 3.47

I2 = 8 - 3.47 = 4.53 A

Hence, I2 = 4.53 A

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