The triangular loop of wire shown in the figure below carries a current I = 5.30
ID: 1264848 • Letter: T
Question
The triangular loop of wire shown in the figure below carries a current I = 5.30 A in the direction shown. The loop is in a uniform magnetic field that has magnitude B = 2.80 T and the same direction as the current in side PQ of the loop
(a) Find the force exerted by the magnetic field on each side of the triangle.
(b) What is the net force on the loop?
N
(c) The loop is pivoted about an axis that lies along side PR. Use the force calculated in part (a) to calculate the torque on each side of the loop.
(d) What is the magnitude of the net torque on the loop?
(f) Calculate the net torque from the torques calculated in part (c) and also from
Do these results agree?
yesno
(e) Is the net torque directed to rotate point Q into the plane of the figure out or out of the plane of the figure?
out of the planeinto the plane
side PQ N side RP N side QR N The triangular loop of wire shown in the figure below carries a current I = 5.30 A in the direction shown. The loop is in a uniform magnetic field that has magnitude B = 2.80 T and the same direction as the current in side PQ of the loop Do these results agree? yesno (e) Is the net torque directed to rotate point Q into the plane of the figure out or out of the plane of the figure? out of the planeinto the plane (a) Find the force exerted by the magnetic field on each side of the triangle. (b) What is the net force on the loop? N (c) The loop is pivoted about an axis that lies along side PR. Use the force calculated in part (a) to calculate the torque on each side of the loop. (d) What is the magnitude of the net torque on the loop? (f) Calculate the net torque from the torques calculated in part (c) and also fromExplanation / Answer
The magnitude of the force exerted by the magnetic field on side PQ of the triangle
= BIL sin 0 = 0
RQ = sqrt (0.62 + 0.82) = 1
Angle between B and I is found from sin theta = 0.8 / 1 = 0.8
The magnitude of the force exerted by the magnetic field on side RQ of the triangle
= BIL sin theta = 2.80 * 5.3 * 1 * 0.8 = 12 N
magnitude of the force exerted by the magnetic field on side PR of the triangle
= BIL sin theta = 2.8 * 5.3 * 0.8 sin 90 = 12 N
Resultant force 12 + 12 = 24 N
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