An electron is accelerated through a potential difference of 10.0 kV. An electro

Question

An electron is accelerated through a potential difference of 10.0 kV.

An electron is accelerated through a potential difference of 10.0 kV. (a) Find its maximum velocity (5.93x10^7ms^-1). It is then passed through a region of uniform magnetic field which has a direction that is perpendicular to the direction of motion. It moves in a circular path of radius 0.10 m in the uniform magnetic field. (b) Find the magnitude of this magnetic field and show its direction on a clearly labelled diagram. (3.38x10^-3T). (c) Find the time taken for the electron to complete one rotation (1.06x10^-8s).

Explanation / Answer

Mass of electron (m) =9.1*10-31kg

charge of an electron(q) =1.6*10-19C

radius(r) =0.10m

Potential difference(V) =10*103V=104V

a)

The maximum velocity is vmax=Sqrt(2*q*V/m)

                                             =5.929*10^7m/s

b) qvB=mv^2/r ====>B=mv/qr =3.38*10-3Tesla

c)

T =2pir/v=(2*3.14*0.10)/5.929*10^7m/s

           =1.06*10-8s

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