A rectangular loop of wire with sides H = 39 cm and W = 60 cm is located in a re

Question

A rectangular loop of wire with sides H = 39 cm and W = 60 cm is located in a region containing a constant magnetic field B = 1.03 T that is aligned with the positive y-axis as shown. The loop carries current I = 343 mA. The plane of the loop is inclined at an angle ? = 23o with respect to the x-axis.

1)

What is ?x, the x-component of the magnetic moment vector of the loop?

xxxx    A-m2

2)

What is ?y, the y-component of the magnetic moment vector of the loop?

xxxx A-m2

3)

What is ?z, the z-component of the torque exerted on the loop?

xxxx N-m

4)

What is Fbc, the magnitude of the force exerted on segment bc of the loop?.

xxx   N

Rectangular Current Loop A rectangular loop of wire with sides H 39 cm and W 60 cm is located in a region containing a constant magnetic field B = 1.03 T that is aligned with the positive y-axis as shown. The loop carries current I = 343 mA. The plane of the loop is inclined at an angle = 23° with respect to the x-axis. W I lan view side view 1) What is fix, the x-component of the magnetic moment vector of the loop? A-m Submit You currently have 0 submissions for this question. Only 5 submission are allowed. You can make 5 more submissions for this question. what is ly, the y-component of the magnetic moment vector of the loop? A-m Submit You currently have 0 submissions for this question. Only 5 submission are allowed. You can make 5 more submissions for this question. 3) What is Tz, the z-component of the torque exerted on the loop? N-m Submit You currently have 0 submissions for this question. Only 5 submission are allowed. You can make 5 more submissions for this question. 4) What is Fbc, the magnitude of the force exerted on segment bc of the loop?. N Submit You currently have 0 submissions for this question. Only 5 submission are allowed. You can make 5 more submissions for this question. 5) What is the direction of the force that is exerted on the segment bc of the loop? along negative x-direction along negative y-direction along postive x-direction along positive y-direction 0 none of the above Submit You currently have 1 submissions for this question. Only 5 submission are allowed. You can make 4 more submissions for this question.

Explanation / Answer

a) M = I*A (along the normal)

So Mx = -0.39*0.60*0.343*sin(23) = 0.031361

b) 0.39*0.60*0.343*cos(23) = 0.073882

c)Torque = M |cross| B

Ty = My By = 0.076098 Nm

d) F = ILB sin(90) = 0.343*0.39*1.03 = 0.1377831 N

I hope this helps.. Please rate.

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