problem29_07

# problem29_07 - 2 NAB0 d B d 2 t 2 t NAB0 1 cos sin for dt...

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29.7: a) for 2 sin 2 2 cos 1 0 0 - = - - = Φ - = T t T NAB T t NAB dt d dt d B π π π ε otherwise. zero ; 0 T t < < . b) 2 at 0 T t = = ε c) . 4 3 and 4 at occurs 2 0 max T t T t T πNAB = = = ε d) From B t T , 0 2 < < is getting larger and points in the z + direction. This gives a
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