An electron in a cathode-ray tube is accelerated through a potential difference

Question

An electron in a cathode-ray tube is accelerated through a potential difference of 10 kV, then passes through the d = 1.6-cm-wide region of uniform magnetic field in the figure(Figure 1) .

An electron in a cathode-ray tube is accelerated through a potential difference of 10 kV, then passes through the d = 1.6-cm-wide region of uniform magnetic field in the figure(Figure 1). What field strength will deflect the electron by ? = 8.0?? Express your answer to two significant figures and include the appropriate units.

Explanation / Answer

from figure

sin(o) =d/r

r =d/sin(o) =1.6/sin8 =11.5 cm

from conservation of energy

qV =(1/2)mv^2

(1.6*10^-19)*(10*10^3)=(1/2)*(9.11*10^-31)*v^2

v=5.93*10^7 m/s

and

r=mv/qB

=>B =mv/qr =(9.11*10^-31)*(5.93*10^7)/(1.6*10^-19)(11.5*10^-2)

B=2.93*10^-3 T or 2.93mT

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