1) What is F ad,x , the x-component of the force exerted by the infinite wire on
ID: 2243335 • Letter: 1
Question
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 = 76 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?
A rectangular loop of wire with sides H = 39 cm and W = 72 cm carries current I2 = 0.37 A. An infinite straight wire, located a distance L = 38 cm from segment ad of the loop as shown, carries current I1 = 0.741 A in the positive y-direction. What is Fad,x, the x-component of the force exerted by the infinite wire on segment ad of the loop? What is Fbc,x, the x-component of the force exerted by the infinite wire on segment bc of the loop?. What is Fnet,y, the y-component of the net force exerted by the infinite wire on the loop? Another infinite straight wire, aligned with the y-axis is now added at a distance 2L = 76 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?Explanation / Answer
1) Fadx = uo*I1*I2*ad/(2*pi*L) = (4*3.14*10^-7*0.741*0.37*0.39)/(2*3.14*0.38) = 56.277e?9 N
2) Fbcx = -uo*I1*I2*bc/(2*pi*(L+w)) = (4*3.14*10^-7*0.741*0.37*0.39)/(2*3.14*1.2) = 17.82105e?9 N
3) Fnet y = 0
4) (uo*I1*I2*ad)/(2*pi*L) = (uo*I2*I*ad)/(2*pi*(2L+w))
I1/L = I/(2L+w)
0.741/0.38 = I/(0.76+0.72)
I = 2.886 along the positive y-direction
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