An electron moves in a circular path perpendicular toaconstant magnetic field of

Question

An electron moves in a circular path perpendicular toaconstant magnetic field of magnitude 1.00 mT. theangularmomentum of the electron about the center of the circle is4.00 x10-25 kg*m2/s. Detrmine (a) The radius of the circular path and (b) The speed of the electron Please explain in detail how equation (1)becomesJ/r2 = Bq ---------------------------------------------------------------- Magnitude of the magnetic field B = 1mT Angular momentum of the electron J = 4 x10-25J*s Charge of the elctron q = 1.6 x 10-19 C

Explanation / Answer

mv / r = Bq ( r/ r) ( mv / r) = Bq ( mvr / r* r) = Bq ( mvr / r 2 ) = Bq ( J / r 2 ) = Bq   Since angularmomentum J = mvr

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