A rectangular loop of wire with sides H = 22 cm and W = 58 cm carries current I

Question

A rectangular loop of wire with sides H = 22 cm and W = 58 cm carries current I2 = 0.393 A. An infinite straight wire, located a distance L = 30 cm from segment ad of the loop as shown, carries current I1 = 0.755 A in the positive y-direction.

1)What is Fad,x, the x-component of the force exerted by the infinite wire on segment ad of the loop?

N

2)What is Fbc,x, the x-component of the force exerted by the infinite wire on segment bc of the loop?.N

3)What is Fnet,y, the y-component of the net force exerted by the infinite wire on the loop?N

4)Another infinite straight wire, aligned with the y-axis is now added at a distance 2L = 60 cm from segment bc of the loop as shown. A current, I3, flows in this wire. The loop now experiences a net force of zero.

What is the direction of I3?

1)along the positive y-direction

2)along the negative y-direction

5)What is the magnitude of I3?

A

Explanation / Answer

here,
H = 22 m
W = 58 m
I1 = 0.755 A
I2 = 0.393 A

Length of wire, L = 30 cm = 0.3 m

Part A :

Force = (Height * uo * I1*I2) / (2 *pi *L)
Where,
uo, magnatic permeability = 1.26 * 10^-6 H
I1, I2 are currents,
L = length of wire

F = (Height * uo * I1*I2) / (2 *pi *L)
F = (22 * 1.26 * 10^-6 * 0.755*0.393) / (2 *3.14 *0.3)
F = 4.36 * 10^-6 N

Part B :

Force = (Height * uo * I1*-I2) / (2 *pi *( L + W ) )

Force = (22 * 1.26 * 10^-6 * 0.755 * -0.393) / (2 *3.14*( 0.30 + 58 ) )

Force = -2.24 * 10^-8 N

Part C :
Think about the possible contributions to Fy from the top and bottom wire and the directions of the
currents

Fnet, y = 0

Part D :

Option 2 is correct.

The current I1 produces a net force on the loop in the positive x-direction. For the current I3 to produce a net force on the loop that cancels the force from I1, it must be directed in the negative y-direction to create a magnetic field in the region of the loop that is directed in the negative z-direction.

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