I\'m not sure where to begin on this. Please help! The accelerating voltage that

Question

I'm not sure where to begin on this. Please help!

The accelerating voltage that is applied to an electron gun is 64 kV, and the horizontal distance from the gun to a viewing screen is 0.54 m.
The charge on an electron is 1.60218 × 10^-19 C and its mass is 9.10939 × 10^-31 kg.
What is the deflection caused by the vertical component of the Earth’s magnetic field or strength 4 × 10^-5 T, assuming that any change in the horizontal component of the beam velocity is negligible.
Answer in units of m.

Thank you!

Explanation / Answer

From Thomson's expariment, the deflection of the electron is given by y = (e*B^2*l^2)/(2*E*m) Where e is the charge of electron = 1.60218*10^-19 C B is the magnetic field = 4*10^-5 T l is the distance of the screen = 0.54 m m is the mass of the electron = 9.10939*10^-31 kg E is the electric field = V/d = 64*10^3 V / 0.54 m = 1.185*10^5 V/m y = [(1.60218*10^-19 C)*(4*10^-5 T)^2*(0.54 m)^2)]/[2*(1.185*10^5 V/m)*(9.10939*10^-31 kg)] y = 3.46*10^-4 m

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