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Class Examples from Chapter 30

# Class Examples from Chapter 30 - Class Examples from...

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Class Examples from Chapter 30 12. To have an induced emf, the magnetic field must be perpendicular (or have a nonzero component perpendicular) to the coil, and must be changing with time. (a) For 2 ˆ (4.00 10 T/m) k B y , / 0 dB dt and hence = 0. (b) None. (c) For 2 ˆ (6.00 10 T/s) k B t , = d Φ B dt = A dB dt = (0.400 m × 0.250 m)(0.0600 T/s) = 6.00 mV, or | | = 6.00 mV. (d) Clockwise. (e) For 2 ˆ (8.00 10 T/m s) k B yt , B = (0.400)(0.0800 t ) ydy = 3 1.00 10 t , in SI units. The induced emf is / 1.00 mV, d B dt    or | | = 1.00 mV. (f) Clockwise. (g) 0 0 B   . (h) None. (i) 0 0 B   . (j) None. 13. The amount of charge is 3 2 2 1 1.20 10 m ( ) [ (0) ( )] [ (0) ( )] [1.60T ( 1.60T)] 13.0 2.95 10 C . B B A q t t B B t R R    14. Figure 30-40(b) demonstrates that / dB dt (the slope of that line) is 0.003 T/s. Thus, in absolute value, Faraday’s law becomes

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( ) B d d BA dB A dt dt dt       where A = 8 ×10 4 m 2 . We related the induced emf to resistance and current using Ohm’s law. The current is estimated from Fig. 30-40(c) to be i = / dq dt = 0.002 A (the slope of that line). Therefore, the resistance of the loop is 4 2 | | | / | (8.0 10 m )(0.0030 T/s) 0.0012 0.0020 A A dB dt R i i . 17. Equation 29-10 gives the field at the center of the large loop with R = 1.00 m and current i ( t ). This is approximately the field throughout the area ( A = 2.00 10 4 m 2 ) enclosed by the small loop. Thus, with B = 0 i /2 R and i ( t ) = i 0 + kt , where i 0 = 200 A and k = ( 200 A 200 A)/1.00 s = 400 A/s, we find (a) 7 4 0 0 4 10 H/m 200A ( 0) 1.26 10 T, 2 2 1.00m i B t R  (b)  7 4 10 H/m 200A 400A/s 0.500s ( 0.500s) 0, 2 1.00m B t  and (c)  7 4 4 10 H/m
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