A loop of wire has the shape of a right triangle (see the drawing) and carries a

Question

A loop of wire has the shape of a right triangle (see the drawing) and carries a current of I = 4.10 A. A uniform magnetic field is directed parallel to side AB and has a magnitude of 1.40 T.

(a) Find the magnitude and direction of the magnetic force exerted on each side of the triangle.

side AC 16.4 N directed into the page

side CB 16.4 N directed out of the page

side BA ___N directed __________

Find side BA and its direction please

(b) Determine the magnitude of the net force exerted on the triangle.

0 N

Explanation / Answer

(a)

For each side, F = iLxB, a vector cross product

|F| = ILBsin, where is the angle between IL and B.

for AC

The angle between the magnetic field and current is q=90o

length of CA = AB tan 55o

FCA = IBL­ sin

FCA = 4.1 * 1.4 * 2 * tan 55o* sin 90o

FCA = 16.4 N, Direction: into page

For CB

The angle between the magnetic field and current is =55o

length of BC = AB/ cos 55o

FBC = IBL sin

FBC = 4.1 * 1.4 *(2/cos 55o) * sin 55o

FBC = 16.4 N, Direction: out of page

For BA

The angle between the magnetic field and current is = 0o

FBA = IBLsin 0o = 0 , No Direction

(b)

Net Force is 0N

FCA = FCB, both are in opposite direction

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