4) In the following velocity selector, the positive plate of the capacitor is ab

Question

4) In the following velocity selector, the positive plate of the capacitor is above the negative as shown. The voltage of the battery is 24 volts and the distance between the plates is 13 milli-meter (no dielectric). A uniform magnetic field of 69 milli-Tesla is in the -z direction. A point charge of -23 micro-coulombs with a mass of 9 grams is passing through at 18 percent of the speed not to be deflected. It is initially moving horizontally in the +x direction. If the length of the plates is 3 meters, how far will the charge be deflected (vertically) in micro-meters (1 x 10-6 m) by the time it exits the velocity selector? Ignore the weight.

Explanation / Answer

Given,

V = 24 V ; d = 13 mm ; B = 69 mT ;

q = -23 uC ; m = 9 g ; v' = 18 v

L = 3 m ;

Let the deflection be s

We know that,

E = V/d = 24/(13 x 10^-3) = 1.85 x 10^3 N/C

Fe = Fb

q E = q v B

v = E/B = 1.85 x 10^3/(69 x 10^-3) = 2.68 x 10^4 m/s

v' = 0.18 x 2.68 x 10^4 = 4.82 x 10^3 m/s

t = d/v = 3/4.82 x 10^3 = 6.22 x 10^-3 s

Now, a = Fnet/m

Fe = q E = 23 x 10^-6 x 1.85 x 10^3 = 0.0425

Fm = 0.18 x 0.043 = 0.0077

Fnet = 0.0425 - 0.0077 = 0.03485 = 0.035

a = 0.035/9 x 10^-3 = 3.89 m/s^2

s = ut + 1/2 at^2

s = 0 + 0.5 x 3.89 x (6.22 x 10^-3)^2 = 75.25 x 10^-6 m

Hence, s = 75.25 micro meter

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