If a proton (m_p = 1.673 Times 10^-27 kg, q_p = e = 1.602 Times 10^-19 C) is acc

Question

If a proton (m_p = 1.673 Times 10^-27 kg, q_p = e = 1.602 Times 10^-19 C) is accelerated through a potential difference of 100 V, What is its velocity? If it passes through a velocity selector and then into a region of space containing only a uniform magnetic field perpendicular to the velocity, What is the strength of that field if the diameter of the proton's orbit is 50 cm? If the magnetic field in the velocity selector is the same as that in part c. What is the strength of the electric field in the velocity selector? If a deuteron (m_d = 2 m_p, q_d = q_p) is to pass through the same velocity selector, through What potential difference must it be accelerated? What then is the diameter of its orbit in the uniform magnetic Held region?

Explanation / Answer

Use the following expression for part a):

1/2 mv2 = eV --> solving for v:

v = (2eV/m)1/2
v = (2*1.602x10-19*100 / 1.673x10-27)1/2
v = 1.38x105 m/s

For part b) the expression to that is the following:

v = E/B

and E = V/d so:

v = V/d*B ---> solving for B:

B = V/d*v

Now in this case we do not know the speed of the velocity selector, but in this case, I will assume the previous velocity calculated:

B = 100 / 0.05 * 1.38x105
B = 0.0145 T

For part c) in the electric field:

E = B*v
E = 0.0145 * 1.38x105 = 2001 N/C

Now I have my doubt in part d) because I don't have the speed of that deuteron, although we have the mass, and we can use the first expression:

1/2 mv2 = eV
V = mv2 / 2e

Now, My doubt is that I don't know if we can use the same value of velocity calculated in part a), but if we can use it, then just solve for V and then for part e) use the equation from part b) and solve instead for d.

Hope this helps

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