Chapter 22 Problems and Answers

Now τ square τ circle bil 2 16bil 2 4 π 125616

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Now τ square / τ circle = BIl 2 /16/BIl 2 /4 π = 12.56/16 = 0.785 22.44 Each of the 10 turns of wire in a vertical, rectangular loopcarries a current of 0.22 A. The loop has a height of 8.0 cm and awidth of 15 cm. A horizontal magnetic field of magnitude 0.050 T isoriented at an angle of relative to the normal to the planeof the loop, as indicated in Figure22–43 . Find (a) the magnetic force on each side of the loop, (b) the net magnetic force on the loop, and (c) themagnetic torque on the loop. (d) If the loop canrotate about a vertical axis with only a small amount of friction,will it end up with an orientation given by θ =0, θ =90 or θ =180? Explain.
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The force F top = N B I Lsin θ = 10(0.22)(0.15)(0.05)sin25 = 6.9732*10 -3 N F bottom = 6.9732*10 -3 N F left = - N B I L sin θ =10(0.22)(0.08)(0.05)sin90 = 8.8*10 -3 N F right = - F left =8.8*10 -3 N b) net force is F net = F top +F bottom + F left +F right = 0 c)torque t = B I A NSin θ =10(0.22)(0.15)(0.15)(0.05)sin65 =1.2*10 -3 N d) The force on the left and right will rotate inclockwiseuntil its normal in the direction with B. It will end up itsorientations at θ = 0 o
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