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# Chapter%2010%20guided%20review%20answers - with a diagram...

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Chapter 10: Guided review answers. Numbers 9 and 10 in the guided review should be reversed. 1. 1.2m or more. 2. It could be the same current, same direction, 20cm from the first. It could also be a current I 2 at a distance r 2 + 10cm from the first current (I 1 ) such that I 2 /r 2 = I 1 /r 1 , where r 1 = 0.1m. 3. B = 0.025T when B is perpendicular to I. Other magnetic fields, parallel to the current, can be added without changing the force. 4. (a) F = 0 on the sides in the z-direction. F = 1.2N on the sides along the y-direction. F on top side is in the x-direction, F on bottom side is in the (–x)-direction. (b) τ = 0.24Nm. 5. The normal to the loop has to be parallel to B. 6. sin θ = 0.5, θ = 30 o , where θ is the angle between B and the normal to the loop,
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Unformatted text preview: with a diagram as on page 306, but turned through 90 o so that B is in the vertical direction. 7. Down. 8. (a) mv = qBr. (b) The angular velocity ω = v/r = qB/m. This is the number of radians per second. The number of revolutions per second is ω /2 π = qB/2 π m, independent of v and r. (c) K = m 2 v 2 /2m = q 2 B 2 r 2 /2m = 1.92MeV. 9. (As marked, goes with Example 10) B = 0.125T. 10. (Goes with example 9.) If we assume that we can use the relation for the infinitely long solenoid, B = 0.314T. 11. (a) Φ = 0 at t = 0, and after rotating 180 o , Φ = 0.18Tm 2 at 90 o and 270 o . (b) Φ = Φ max sin θ , E =-E max cos θ , with diagrams as on page 322, with T = 1/12 s....
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