A rectangular loop of wire with sides H- 32 cm and W 71 cm carries current I2 0.

Question

A rectangular loop of wire with sides H- 32 cm and W 71 cm carries current I2 0.314 A. An infinite straight wire, located a distance L- 32 cm from segment ad of the loop as shown, carries current 0.707 A in the positive y-direction. 8) What is Fadx, the x-component of the force exerted by the infinite wire on segment ad of the loop? 9) What is Fbcx, the x-component of the force exerted by the infinite wire on segment bc of the loop?. 10) What is Fnet,y, the y-component of the net force exerted by the infinite wire on the loop? 11) 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 2L aW I What is the direction of I3? a. along the positive y-direction b. along the negative y-direction 12) What is the magnitude of I3?

Explanation / Answer

8)

We know that magnetic field dure to an infinte straight current carrying conductor is

B=I/2x....

Where I is the current is in wire and x is the distance at which we are calculating the magnetic field.

Now, we know that force on a current carrying conductor placed in a magnetic field B is

F=i(dl*B)

Now dl is in -y direction

and B is in the plane perpendicular to the plane of paper and pointing inside.

So F will be in poistive x direction.

So,

F=I2(H*I1/2L)

F=0.314*(0.32*2*10-7*0.707/0.32)

F=4.43996*10-8N

9)

Similarly for branch bc, only x will be L+W.

So,

F=0.314*(0.32*2*10-7*0.707/0.32+0.71)

F=1.379*10-8N and it will be in the negative x direction as dl is in positive y direction.

10)

Fnet in y direction will be zero as branch dc and ab are compeletly symmetrical and dl is in opposite direction for each of them, so they will cancel out each other's magnetic field.

11)

direction of I3 will be in positive y-direction as net force on the loop due to I1 is in the positive x direction , so to make net force zero, force due to I3 should be in the negative x-direction.

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