Show exact answers and work, thanks! Will rate for correct answer! The triangula
ID: 1283238 • Letter: S
Question
Show exact answers and work, thanks! Will rate for correct answer!
The triangular loop of wire shown in the drawing carries a current of I = 5.34 A. A uniform magnetic field is directed parallel to side AB of the triangle and has a magnitude of 1.94 T.(a) Find the magnitude of the magnetic force exerted on each side of the triangle. (b) Determine the magnitude of the net
force exerted on the triangle.
Explanation / Answer
FAB = B I L sin 180
= 1.94 * 5.34 * 2.00 * sin 180
FAB = 0
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The length of AC is tan theta = AC / AB
AC = AB tan theta
AC = (2.00 m) tan 55
FAC = BI (AC) sin 90
= 1.94 * 5.34 * (2.00 m) tan 55 * sin 90
FAC = 29.6 N [direction of force is into the page]
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length of BC
cos theta = AB / BC = AB / cos 55 = 2 / cos 55 = 3.49 m
FBC = BI (BC) sin 90
= 1.94 * 5.34 * 3.49 * sin 55
FAC = 29.6 N [direction of force is out the page]
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b)Fnet = FAB + FBC + FAC
= 0 + 29.6 - 29.6
Fnet = 0