A proton with a speed of 1.8 x 106 m/s is shot into a region between two plates
ID: 2054645 • Letter: A
Question
A proton with a speed of 1.8 x 106 m/s is shot into a region between two plates that are separated by a distance of 0.26 m. As the drawing shows, a magnetic field exists between the plates, and it is perpendicular to the velocity of the proton. What must be the magnitude of the magnetic field, so the proton just misses colliding with the opposite plate?
A proton with a speed of 1.8 x 106 m/s is shot into a region between two plates that are separated by a distance of 0.26 m. As the drawing shows, a magnetic field exists between the plates, and it is perpendicular to the velocity of the proton. What must be the magnitude of the magnetic field, so the proton just misses colliding with the opposite plate?Explanation / Answer
We know that the proton cannot hit the other plate, which means the radius of it's circular path cannot be larger than 0.26 m.
If we use the equation:
r = mv/qB
Where r is the radius of it's circular path, m is the mass of the particle, q is the charge of the particle and B is the magnetic field, we simply rearrange for B, the magnetic field:
B = mv/qr
We now plug in our known values:
B = [(1.8 * 10^6 m/s)(1.67 x 10-27kg)] / [(1.602 x 10-19C)(0.26m)]
Where mp = 1.67 x 10-27kg and the charge of a proton is q = 1.602 x 10-19C
We now solve:
B = 7.21*10^-2 T
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