The two 5.0-cm-long parallel electrodes in the figure (Figure 1) are spaced 1.0
ID: 1503551 • Letter: T
Question
The two 5.0-cm-long parallel electrodes in the figure (Figure 1) are spaced 1.0 cm apart. A proton enters the plates from one end. an equal distance from both electrodes. A potential difference Delta V = 500 V across the electrodes deflects the proton so that it strikes the outer end of the lower electrode. What magnetic field strength will allow the proton to pass through undeflected while the 500 V potential difference is applied? Assume that both the electric and magnetic fields are confined to the space between the electrodes. Express your answer to two significant figures and include the appropriate units. What magnetic field direction will allow the proton to pass through undeflected while the 500 V potential difference is applied? upward downward to the left to the right into the page out of the pageExplanation / Answer
E = V/d
= 500/(0.01) = 50000 V/m
F = q E
= 1.6e-19 * 50000 = 8e-15 N
a = F/m
= 8e-15/(1.67e-27) = 4.79042e12 m/s2
y = 0.5 a t^2
==> t = sqrt(2y/a)
= sqrt(2*(0.01/2)/(4.79042e12)) = 4.569e-8 s
==> v = x/t
= 0.05/4.569e-8 = 1.0943e6 m/s
magnetic field:
B = E/v
= 50000/1.0943e6 = 0.0457 T
According to the right hand rule, the magnetic field is into the page
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