A velocity selector is built with a magnetic field of magnitude 5.2 T and an electric field of strength 4.6 × 10 ^4 N/C. The directions of the two fields are perpendicular to each other. At what speed will electrons pass through the selector without deflection? Let the charge of an electron q = −1.6 × 10 ^ −19
A.8.8 × 10^3 m/s
B 2.4 × 10 ^ - 15 m/s
C 7.9 × 10 ^ 13 m/s
D 2.0 x 10 ^ 4 m/s
q E = q V B
V = E/B for any old q
= 4.6*10^4 / 5.2
set the two forces equal.
Eq =q vB
v= E/B=4.6e4/(5.2)= ..... m/second
that's not an answer choice.
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To determine the speed at which electrons pass through the velocity selector without deflection, we need to find the equilibrium between the electric force and the magnetic force acting on the electron.
The electric force on an electron is given by Fe = q * E, where q is the charge of an electron and E is the electric field strength. In this case, q = -1.6 × 10^-19 C and E = 4.6 × 10^4 N/C.
The magnetic force on an electron is given by Fm = q * v * B, where v is the velocity of the electron and B is the magnetic field strength. In this case, q = -1.6 × 10^-19 C, B = 5.2 T.
In equilibrium, the electric force and the magnetic force are equal, so we have:
q * E = q * v * B
Simplifying and solving for v:
E = v * B
v = E / B
Plugging in the values:
v = (4.6 × 10^4 N/C) / (5.2 T)
Now, we need to convert N/C to V/m and T to N/(C·m/s). The conversion is 1 N/C = 1 V/m and 1 T = 1 N/(C·m/s). So:
v = (4.6 × 10^4 V/m) / (5.2 N/(C·m/s))
Simplifying further, we can multiply the numerator and denominator by s/C to get:
v = (4.6 × 10^4 V/m) * (s / C) / (5.2 N/(C·m/s))
Cancelling out the C and rearranging the terms, we have:
v = (4.6 × 10^4 V/m) * (s / 5.2 N) * (m/s)
Now, the units V/m, N, and m/s cancel out, giving us the final answer in meters per second:
v = (4.6 × 10^4 / 5.2) m/s
Evaluating this expression gives us approximately:
v ≈ 8.8 × 10^3 m/s
So the correct option is A) 8.8 × 10^3 m/s.