Class examples from Chapter 32

Class examples from Chapter 32 - Class examples from...

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1 Class examples from Chapter 32 2. (a) The flux through the top is +(0.30 T) r 2 where r = 0.020 m. The flux through the bottom is +0.70 mWb as given in the problem statement. Since the net flux must be zero then the flux through the sides must be negative and exactly cancel the total of the previously mentioned fluxes. Thus (in magnitude) the flux though the sides is 1.1 mWb. (b) The fact that it is negative means it is inward. 6. The integral of the field along the indicated path is, by Eq. 32-18 and Eq. 32-19, equal to 00 2 enclosed area (4.0 cm)(2.0 cm) (0.75 A) 52 nT m total area 12 cm d i     . 7. (a) Inside we have (by Eq. 32-16) 2 01 /2 d B i r R  , where 1 0.0200 m, r 0.0300 m, R and the displacement current is given by Eq. 32-38 (in SI units): 12 2 2 3 14 0 (8.85 10 C /N m )(3.00 10 V/m s) 2.66 10 A E d d i dt . Thus we find 7 14 19 22 (4 10 T m/A)(2.66 10 A)(0.0200 m) 1.18 10 T 2 2 (0.0300 m) d ir B R   .
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Class examples from Chapter 32 - Class examples from...

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