An electron in a cathode ray tube accelerates across apotential difference 10.0

Question

An electron in a cathode ray tube accelerates across apotential difference 10.0 kv then passes through a small aperture.In one of the electrodes as it emerges on the other side of thiselectrodevit immediately enters a region in which a uniformmagnetic field of width 2.0 cm is present. As the electron reachesthe fat edge of the magnetic field its path is deflected 20 degreesfrom the horizontal direction along which is initially startedmoving. Determine the velocity, v of the electron and thetime elapsed, t. Step by step solutions please... An electron in a cathode ray tube accelerates across apotential difference 10.0 kv then passes through a small aperture.In one of the electrodes as it emerges on the other side of thiselectrodevit immediately enters a region in which a uniformmagnetic field of width 2.0 cm is present. As the electron reachesthe fat edge of the magnetic field its path is deflected 20 degreesfrom the horizontal direction along which is initially startedmoving. Determine the velocity, v of the electron and thetime elapsed, t. Step by step solutions please...

Explanation / Answer

V = 10.0 kV, d = 2.0 cm, = 20o, find v andt energy relation: qV = mv2/2 v = (2qV/m) = 5.93*107 m/s let radius = r when it is in the magnetic field d = rsin, so r = d/sin during period T = 2r/v it turns 360o, so it takest = T/360* to turn t = 2r/v * /360 =d/(360vsin) = 1.72*10-10 s

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