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Q. A beam of electrons travelling with a speed of 4 X 10⁷ ms^-1 is projected normal to a magnetic field of B = 10^-3 Wb m^-2. What is the path of the beam in the magnetic field?

Solution: Since the electrons are released usually to the magnetic field the electrons travel in a circular path.

∴ Bev = mv²/ r

r = mv/Be = 9.11 X 10^-31X 4 X 10⁷ / 10^-3 X 1.6 X 10^-19

Radius of the path r = 0.2275 m.

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Category:- Physics
Reference No.:- M9571541

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