A rectangular loop of wire with sides H = 24 cm and W = 74 cm carries current I

Question

A rectangular loop of wire with sides H = 24 cm and W = 74 cm carries current I2 = 0.165 A. An infinite straight wire, located a distance L = 22 cm from segment ad of the loop as shown, carries current I1 = 0.768 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 = 44 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

Your submissions:

B

Submitted:

Sunday, March 26 at 3:40 PM

Feedback:

Your answer 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.

5)

What is the magnitude of I3?

A

Explanation / Answer

Q1.

direction of current in ad of the loop and and the infinite wire are in different direction.

so force is repulsive in nature.
hence force is along +ve x axis.

force magnitude=mu*I1*I2*length of ad/(2*pi*distance)

=4*pi*10^(-7)*0.768*0.165/(2*pi*0.22)

=1.152*10^(-7) N

Q2.

direction of current in bc and the infinite wire is along the same direction

hence force is attractive in nature

force magnitude=mu*I1*I2*length of bc/(2*pi*distance)

=4*pi*10^(-7)*0.768*0.165/(2*pi*(0.22+0.74))

=2.64*10^(-8) N

as force is along -ve x axis,

answer=-2.64*10^(-8) N

Q3.

total force is along x axis.

so Fnet,y=0

Q4.

as I3 is closer to segment bc, force magnitude on bc would be greater as compared to force magnitude on ad.

due to the I1, force is along +ve x axis.

in order to make total force to be zero,

I3 be such that force due to I3 on the segment bc is along -ve x axis.

so force is repulsive in nature.

hence current I3 and current in bc are along different direction.

so I3 will be in downward direction

correct answer: along negative y direction

Q5.

total force along -ve x axis due to I3

=force on segment bc - force on segment ad
=(mu*I3*I2*H/(2*pi*2*L)) - (mu*I3*I2*H/(2*pi*(2*L+W))

=(4*pi*10^(-7)*I3*0.165*0.24/(2*pi))*((1/0.44)-(1/1.18))

=1.1288*10^(-8)*I3

this force magnitude should be equal to total force magnitude due to I1.

hence 1.1288*10^(-8)*I3=1.152*10^(-7)-2.64*10^(-8)

==>I3=7.8667 A

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