A rectangular loop of wire with sides H = 29 cm and W = 60 cm carries current I2

Question

A rectangular loop of wire with sides H = 29 cm and W = 60 cm carries current I2 = 0.205 A. An infinite straight wire, located a distance L = 36 cm from segment ad of the loop as shown, carries current I1 = 0.792 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?

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

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

Explanation / Answer

I am assuming that the current in the infinite wire is parallel to the loop abcd.

1) The magnetic field due to the infinte wire at a distance L

Let I1 be the current in the infinte wire = 0.792 A

B = uoI1/2piL =4 × 107 * 0.792 / 2pi*0.36 = 4.4*107 T

To find the force on ad, use:

F = I2Lad × B

Lad = lenght of ad = 0.29 m

I2 be the current in loop = 0.205 A

F = 0.205*0.29*4.4*107 = 26.15*109 N

2) For bc

L = 36+60 = 96 cm =0.96 m

B = uoI1/2piL =4 × 107 * 0.792 / 2pi*0.96 = 1.65*107 T

F = I2Lbc × B

I2 be the current in loop = 0.205 A

Lbc = lenght of bc = 0.29 m

F = -0.205*0.29*1.65*107  = -9.8*109 N

3) The force between two perpendicular current carrying wires is zero. Only the sides of the square parallel to the inifinite wire contribute to the force.

Net Force = 26.15*109 - 9.8*109 = 16.35*109 N

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