(a) Derive an expression for the speed v of a beam of charged particles if, when

Question

(a) Derive an expression for the speed v of a beam of charged particles if, when moving through perpendicular electric and magnetic fields, the beam moves in a straight line at constant velocity directed to the east. The magnetic field is directed north, and the elec tric field is directed vertically downward (in other words, toward the center of the earth). (b) Evaluate the expression for v if E " 3.00 X 10^4 N/C and B =2.00 X 10^-2 T. (c) What would happen to charges moving faster than v? (d) What would happen to charges moving slower than v? (Crossed electric and magnetic fields serve as a velocity selector and are used to produce a beam that has a well-defined velocity as the beam enters a mass spectrometer.)

Explanation / Answer

a) The direction of magnetic force will be vertically upwards and electric force will be vertically downwards, considering a positive charge,

qvB=qE

v=E/B

b) v= 30000/0.02

v=1500000 m/s

c) if the charge moves faster than v then the magnitude of electric force remains constant but the magnitude of magnetic force increases causing the charge to moves vertically upwards instead of going straight.

d) same reason as above, the charge will.move downwards.

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