The four wires in the figure are tilted at 20 with respect to a uniform 0.35 fie

Question

The four wires in the figure are tilted at 20 with respect to a uniform 0.35 field. Each carries 4.5 and is 0.35 long.

A.What is the magnitude of the force on the wire a?

B.What is the direction of the force on the wire a?

C.What is the magnitude of the force on the wire b?

D.What is the direction of the force on the wire b?

E.What is the magnitude of the force on the wire c?

F.What is the direction of the force on the wire c?

G.What is the magnitude of the force on the wire d?

H.What is the direction of the force on the wire d?

Explanation / Answer

F = IL x B = BIL*sin {theta} A.) F = 0.35 * 4.5 * 0.35 * sin 20 = 0.188 N B ) Direction : INTO THE PAGE or PLANE C) F = 0.35 * 4.5 * 0.35 * sin (pi - 20) = 0.188 N D) Direction :INTO THE PLANE or PAGE E) F = 0.35 * 4.5 * 0.35 * sin (pi + 20) = - 0.188 N So, magnitude = 0.188 N F) Direction : Out of the Plane G) F = 0.35 * 4.5 * 0.35 * sin (2pi - 20) = -0.188 N So, magnitude = 0.188 N H) Direction : Out of the Plane

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