An electron enters perpendicularly into a uniform magnetic field which has a flux density of 0.12 T This results in a magnetic force of 9.6 x 10\(^{-2}\) N ...
Question 1 Report
An electron enters perpendicularly into a uniform magnetic field which has a flux density of 0.12 T This results in a magnetic force of 9.6 x 10\(^{-2}\) N on the electron. Calculate the speed of the electron as it enters the magnetic field. (e =1.6 x \(10^{19}\) C)
Answer Details
An electron moving perpendicular to a uniform magnetic field experiences a force
\[F=Bev\sin\theta,\qquad \theta=90^\circ,\ \sin 90^\circ=1.\]Making the speed the subject:
\[v=\frac{F}{Be}.\]Substituting \(F=9.6\times10^{-12}\text{ N}\), \(B=0.12\text{ T}\) and \(e=1.6\times10^{-19}\text{ C}\):
\[v=\frac{9.6\times10^{-12}}{0.12\times1.6\times10^{-19}}=\frac{9.6\times10^{-12}}{1.92\times10^{-20}}=5.0\times10^{8}\text{ m s}^{-1}.\]The speed of the electron as it enters the field is \(5.0\times10^{8}\text{ m s}^{-1}\).
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