A rectangular loop of wire with sides H = 20 cm and W = 53 cm carries current I

Question

A rectangular loop of wire with sides H = 20 cm and W = 53 cm carries current I2 = 0.211 A. An infinite straight wire, located a distance L = 32 cm from segment ad of the loop as shown, carries current I1 = 0.645 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 = 64 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?

along the positive y-direction

along the negative y-direction

5)

What is the magnitude of I3?

A

What is the direction of I3? along the positive y-direction along the negative y-direction 5) What is the magnitude of I3? A A rectangular loop of wire with sides H = 20 cm and W = 53 cm carries current I2 = 0.211 A. An infinite straight wire, located a distance L = 32 cm from segment ad of the loop as shown, carries current I1 = 0.645 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 = 64 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.

Explanation / Answer

Magnetic field due to current carring wire, B = uI/2pia

Magnetic field on AD, Bad = 2x10-7 x 0.645/0.32 = 4.03x10-7 T (Inside the plane of paper)

Force on the wire AD, Fad = Bil = 4.03x10-7 x 0.211 x 0.2 = 1.7 x 10-8 N (Towards + X axis)

Magnetic field on BC, Bbc = 2x10-7 x 0.645/(0.32+0.53) = 1.52 x10-7 T (Inside the plane of paper)

Force on the wire AD, Fad = Bil = 1.52x10-7 x 0.211 x 0.2 = 0.64 x 10-8 N (Towards - X axis)

Force on the AB abd CD will be equal in magnitude but opposite in direction. So, Y component of the force on the loop will be 0.

To mage total force on x direction 0, Force on AD will be towards -X axis so that total force on loop due to I3 towards -X axis. So, its magnetic field should direct inside the loop which results in the I3 direction towards -Y direction.

F = - Bbc i2 l + Bad i2 l

= (u I3 /2pi 2L) i2 l + (u I3/2pi(2L+W)) i2 l

= (u I3/ 2pi) (i2 l) (1/2L + 1/(2L +W))

(1.7-0.64) x10-8 = i3 x 2x10-7 x 0.211 x 0.2 (1/.64 + 1/1.17)

0.106 = I3 x 0.204

I3 = 0.52 A

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