65. (a) The energy delivered by the battery is the integral of Eq. 28-14 (where we use Eq. 30-41 for the current): ()()22battery0 06.702.00 s 5.50 H2115.50H110.0V2.00 s6.706.7018.7 J.ttRt LRt LLPdtedtteRRReεε−−−Ω⎡⎤=−=+−⎢⎥⎣⎦⎡⎤−⎢⎥=+⎢⎥ΩΩ⎣⎦=∫∫(b) The energy stored in the magnetic field is given by Eq. 30-49: ()22226.702.00 s 5.50 H
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