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Chapter%2029%20Solutions%20sp%2008%20p260 - ELECTROMAGNETIC...

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E LECTROMAGNETIC I NDUCTION 29.7.I DENTIFY : Calculate the flux through the loop and apply Faraday’s law. S ET U P : To find the total flux integrate B d over the width of the loop. The magnetic field of a long straight wire, at distance r from the wire, is 0 2 I B r . The direction of B is given by the right-hand rule. E XECUTE : (a) When 0 2 i B r , into the page. (b) 0 . 2 B i d BdA Ldr r (c) 0 0 ln( / ). 2 2 b b B B a a iL dr iL d b a r (d) 0 ln( ) . 2 B d L di b a dt dt E (e) 7 0 (0.240 m) ln(0.360/0.120)(9.60 A/s) 5.06 10 V. 2 E E VALUATE : The induced emf is proportional to the rate at which the current in the long straight wire is changing 29.9.I DENTIFY and S ET U P : Use Faraday’s law to calculate the emf (magnitude and direction). The direction of the induced current is the same as the direction of the emf. The flux changes because the area of the loop is changing; relate dA / dt to dc / dt , where c is the circumference of the loop. (a) E XECUTE : 2 2 2 and so /4 c r A r A c 2 ( /4 ) B BA B c 2 B d B dc c dt dt E At 9.0 s, 1.650 m (9.0 s)(0.120 m/s) 0.570 m t c (0.500 T)(1/2 )(0.570 m)(0.120 m/s) 5.44 mV E (b) S ET U P : The loop and magnetic field are sketched in Figure 29.9. Take into the page to be the positive direction for . A Then the magnetic flux is positive. Figure 29.9 E XECUTE : The positive flux is decreasing in magnitude; / B d dt is negative and E is positive. By the right- hand rule, for A into the page, positive E is clockwise. E VALUATE : Even though the circumference is changing at a constant rate, / dA dt is not constant and E is not constant. Flux is decreasing so the flux of the induced current is and this means that I is clockwise, which checks. 29
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29.16. I DENTIFY : By Lenz’s law, the induced current flows to oppose the flux change that caus ed it. S ET U P and E XECUTE : The magnetic field is outward through the round coil and is decreasing, so the magnetic field due to the induced current must also point outward to oppose this decrease. Therefore the induced current is
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