Part C) Find the magnetic torque on the loop. Part D) If the loop can rotate abo

Question

Part C) Find the magnetic torque on the loop.

Part D) If the loop can rotate about a vertical axis with only a small amount of friction, will it end up with an orientation given by ?=0, ?=90?, or ?=180??

Part A IP Each of the 10 turns of wire in a vertical, rectangular loop carries a current of 0.25 A. The loop has a height of 8.0 cm and a width of 15 cm. A horizontal magnetic field of magnitude 5.3x102 T is oriented at an angle of e 65 relative to the normal to the plane of the loop as indicated in the figure(Figure 1). Find the magnitude of the magnetic force on each side of the loop. Express your answers using two significant figures separated by commas. undo Fop, Flottorght Submit RequestAnswer Part B Figure 1 of 1 Find the net magnetic force on the loop Express your answer using two significant figures. 15 cm 8.0 cm Submit Re st Answer Normal to loop Part C

Explanation / Answer

Given,

N = 10 ; I = 0.25 A ; h = 8 cm ; w = 15 cm ; B = 5.3 x 10^-2 T ; theta = 65 deg

A)we know that, the magnetic force on a wire is:

F = N B I L sin(theta)

Ftop = 10 x 5.3 x 10^-2 x 0.25 x 0.15 sin(90 - 65) = 0.0084 N

Fbottom = -Ftop = -0.0084 N

Fleft = 10 x 5.3 x 10^-2 x 0.25 x 0.08 x sin90 = 0.0106 N

Fright = -Fleft = -0.0106 N

Hence, Ftop = 0.0084 N ; Fbottom = -0.0084 N ; Fleft = 0.0106 N ; Fright = -0.0106 N

b)Net force is

Fnet = Ftop + Fbottom + Fright + Fleft = 0 N

Hence, Fnet = 0 N

c)Net torque

Tau = N B I A sin(theta)

Tau = 10 x 5.3 x 10^-2 x 0.25 x 0.08 x 0.15 x sin65 = 0.00144 N-m

Hence, Tau = 0.00144 N-m

d)when theta = 0, it will align with and tau = 0.

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