The triangular loop of wire shown in the figure below carries a current I = 4.20
ID: 1264951 • Letter: T
Question
The triangular loop of wire shown in the figure below carries a current I = 4.20 A in the direction shown. The loop is in a uniform magnetic field that has magnitude B = 3.00 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. side PQ 0 side RP 10.08 side QR 10.08 (b) What is the net force on the loop? 0 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. side PQ 0 N.m side RP 0 N.m side QR 3.5 x N.m (d) What is the magnitude of the net torque on the loop? 0 XExplanation / Answer
a)
F_PQ = B*I*L*sin(0)
= 0
F_RP = B*I*L*sin(90)
= 3*4.2*0.8
= 10.08 N
F_QR = B*I*L*sin(theta)
= 3*4.2*0.8
= 10.08 N
b) Fnet = 0
c)
T_PQ = 0
T_RP = 0
T_QR = F_QR*r*sin(90)
= 10.08*0.3*1 (here r is perpendicular distance between axis of rotation anf force)
= 3.024 N.m
d) Tnet = 3.024 N.m
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