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# hw08 - PHYS-1200 PHYSICS II HOMEWORK SOLUTIONS CLASS 8...

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PHYS-1200 PHYSICS II SPRING 2005 HOMEWORK SOLUTIONS CLASS 8. 30.02, 30.03, 30.06, 30.08, 30.22 30.02. a) mV 0 . 7 0 . 12 mWb] ) 0 . 7 0 . 6 [( 2 + = + = Φ = t t t dt d dt d B E At t = 2.0 s, 0 . 7 ) 0 . 2 ( 0 . 12 + = E mV E = 31 mV b) The flux is increasing, so Lenz’s law tells us that the field produced by the induced current must point into the page to oppose the increase. The current must flow clockwise around the circuit, or: to the left in resistor R . 30.03. dt dB r dt dB r dt d r B BA dt d B B B 2 2 2 and , so , where , π π π - = = Φ = = Φ Φ - = E E mV 11 Then, s. / T 25 . 0 s 0 . 4 s 0 . 6 T 5 . 0 T 0 . 0 s, 6 < < s 4 For c) mV 0 . 0 Then, s. / T 0 . 0 s 0 . 2 s 0 . 4 T 5 . 0 T 5 . 0 s, 4 < < s 2 For b) mV 11 Then, s. / T 25 . 0 s 0 . 0 s 0 . 2 T 0 . 0 T 5 . 0 s, 2 < < 0 For a) T) / s mV 45 ( T) / s V 045 . 0 ( ) m 12 . 0 ( Then, 2 = - = - - = = = - - = - = = - - = - = - = - = E E E E dt dB t dt dB t dt dB t dt dB dt dB dt dB π 30.06. dt dr rB dt dr B dt r d B dt dA B dt BA d dt d B π π π 2 ) ( ) ( 2 2 - = - = - = - = - = Φ - = E m/s) 0.750 T)( m)(0.800 120 . 0 ( 2 2 - - = - = π π dt dr rB E E = 0.452 V

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30.08. The field (due to the current in the straight wire) is out-of-the-page in the upper half of the circle and is into the page in the lower half of the circle. Due to the symmetry, the magnitude of the flux in the upper half equals that in the lower half producing zero net flux, at any time. There is no induced current in the circle. i = 0 30.22. a) we so position, of function a is loop. gh the flux throu net
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hw08 - PHYS-1200 PHYSICS II HOMEWORK SOLUTIONS CLASS 8...

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