(c) top segment magnitude N direction ---Select--- +x-direction ?x-direction +y-
ID: 2034364 • Letter: #
Question
(c)
top segment
magnitude N direction ---Select--- +x-direction ?x-direction +y-direction ?y-direction +z-direction ?z-direction The magnitude is zero.
(d)
bottom segment
magnitude N direction ---Select--- +x-direction ?x-direction +y-direction ?y-direction +z-direction ?z-direction The magnitude is zero.
(e)
Find the magnitude of the net torque on the loop about the axle.
N · m
A current-carrying rectangular wire loop with width a = 0.125 m and length b = 0.220 m is in the xy-plane, supported by a nonconducting, frictionless axle of negligible weight. A current of I = 2.50 A travels counterclockwise in the circuit (see the figure below). Calculate the magnitude and direction of the force exerted on the left, right, top, and bottom segments of wire (in N) by a uniform magnetic field of 0.350 T that points in the positive x-direction. Find the magnitude of the net torque (in N·m) on the loop about the axle. axle axle HINT left segment of the wire magnitude direction (a) ---Select- (b) right segment of the wire magnitude direction ---Select- (c) top segmentExplanation / Answer
magnetic force F = I(L x B )
(a) magnetic field B = 0.35 i
length L = - bj = -0.22 j
magnetic force F = 2.5*(-0.22j x 0.35 i)
magnetic force F = 0.193 N ( - j x i ) j X i = -k
magnetic force F = 0.193 N K
magnitude = 0.193 N
direction +z direction
----------------------------
(b) magnetic field B = 0.35 i
length L = bj = 0.22 j
magnetic force F = 2.50*(0.22 j x 0.35 i)
magnetic force F = 0.193 N ( j x i ) j X i = -k
magnetic force F = 0.193 N -K
magnitude = 0.193 N
direction -z direction
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(c) magnetic field B = 0.35 i
length L = -ai = 0.125 -i
magnetic force F = 2.5*(0.125 -i x 0.35 i)
magnetic force F = 0.169 N ( -i x i ) i X i = 0
magnetic force F = 0 N
magnitude = 0 N
direction no direction
-----------------------------
(d) magnetic field B = 0.35 i
length L = ai = 0.115 i
magnetic force F = 2.5*(0.115 i x 0.35 i)
magnetic force F = 0.169 N ( i x i ) i X i = 0
magnetic force F = 0 N
magnitude = 0 N
direction no direction
-----------------------------
torque = I*A*B = I*a*b*B
torque = 2.5*0.125*0.22*0.35 = 0.024 Nm
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