Question

Two insulated wires, each 4.42 m long, are taped together to form a two-wire unit that is 4.42 m long. One wire carries a current of 7.33 A; the other carries a smaller current I in the opposite direction. The two-wire unit is placed at an angle of 69.4o relative to a magnetic field whose magnitude is 0.547 T. The magnitude of the net magnetic force experienced by the two-wire unit is 2.24 N. What is the current I

Answer 1

Answer:

Current I=6.34A

Explanation:

Given data

L=L₁=L₂=4.42 m

I₁=7.33A

Angle α=69.4°

B=0.547T

Force F=2.24N

Required

Current I

Solution

The length of each wire ,the magnetic field B,and the angle are same for both wires.

As we know that Force is:

$F_{net}=I_{1}LBSin\alpha -I_{2}LBSin\alpha\\F_{net}=(I_{1}-I_{2})LBSin\alpha\\I_{2}=I_{1}-\frac{F_{net}}{LBSin\alpha}\\I_{2}=7.33A-\frac{2.24N}{(4.42m)(0.547T)Sin(69.4)} \\I_{2}=6.34A$    

Current I=6.34A