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Question

An electron moves in a circular pathperpendicular to a constant magnetic field witha magnitude of 1.0mT . If the angularmomentum of the electron about the centre ofthe circte is 8.1×1038 J.s, determine the speedof the electron. (Take me=9×1031kg) .

A
2 m/s
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B
4 m/s
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C
6 m/s
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D
8 m/s
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Solution

The correct option is B 4 m/s
mvr=8.1×1038r=mvqBmv.mvqB=8.1×1038m2v2=1.6×1019×1×103×8.1×1038v2=1.6×1019×103×8.1×10389×9×1062v=4m/s
Option B is correct .

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