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 = 2.5-cm-wide region of uniform magnetic field in the figure.

What field strength will deflect the electron by ? = 11

An electron in a cathode-ray tube is accelerated through a potential difference of 10 kV, then passes through the d = 2.5-cm-wide region of uniform magnetic field in the figure. What field strength will deflect the electron by ? = 11°? Express your answer to two significant figures and include the appropriate units.

Explanation / Answer

First find v, using the equation K=1/2 mv^2, so V= ?2K/m, but because this is given in Electron Volts (eV) V=?2eK/m where e=1.602x10^-19 the charge of an electron! v=5.93x10^7 m/s.

So now we need to find the Acceleration, and because it is not a straight line we use the Centripetal Acceleration formula a=v^2/r where r is the radius in METERS! So a= (5.93x10^7)^2/0.01m= 3.52x10^17 m/s^2.

Now we have the Velocity and Acceleration, we can find the Magnetic Field (B) by substituting F=ma into F=qvB so ma=qvB which equates to B=ma/qv (m is the mass of an electron, and q is the charge of the electron so it also could be written as B=ma/ev!)

B=(9.109x10^-31)(3.52x10^17) / (1.602x10^-19)(5.93x10^7) = 0.0338. BUT we now have to Multiply this answer by (?). Now because this is a small angle, we can use either sin or tan because of small angle approximation! Im going to use tan.

So 0.0338tan(10)=5.95x10^-3 T

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