A large velocity selector contains an electric ﬁeld E = 70 N/C pointing
down toward the ground and a magnetic ﬁeld B = 7 T pointing parallel to
the ground and into the page of the ﬁgure shown below.
(Reference Attached Image)
A particle having positive charge q = +1 mC and mass m = 5 grams
moves horizontally through this velocity selector along the dashed line in the
ﬁgure. Determine the speed of the charged particle after it passes through
the velocity selector. As usual, you may assume that g = 9.8 m/s^2.
(a) 3 m/s (b) 8 m/s (c) 10 m/s
(d) 13 m/s (e) 17 m/s (f) None of the above.
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The particle has a charge q = 1 mC and mass 5 g. When it moves through the velocity selector there is a force exerted on the particle by the magnetic field, the electric field as well as the gravitational field.
The force exerted by the electric field is given by F = q*E. Here, it is equal to 70*1*10^-3 = 0.07 N in the downward direction. The force due to the magnetic field is given by F = charge*velocity*magnetic field. Here it is equal to 10^-3*v*7 = 0.007*v vertically upwards. The gravitational force acting on the particle is equal to 9.8*5*10^-3 = 0.049 N acting vertically downwards.
The particle continues moving in the horizontal direction from left to right. This is the case when the net vertical forces are equal to 0.
-0.007*v + 0.07 + 0.049 = 0
=> v = 17 m/s
The velocity of the charged particle is 17 m/s
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