using the above diagram. A proton is launched vertically at a night h through a
ID: 1582720 • Letter: U
Question
using the above diagram.
A proton is launched vertically at a night h through a potential difference V and through a slit. above the slit there is a region of width d with no magnetic field.
1. what is q' speed immediately before entering the magnetic field.
2. what is the direction of the magnetic force on q immediately after it enters the magnetic field.
3. viewed from the side of the above diagram,describe in detail the motion of the proton while it is in the magnetic field.
4. what is the greatest height reached by the proton,as measured from the ground.
5. does the proton ever leave the magnetic field?
Explanation / Answer
1.
using conservation of energy
energy gained due to loss of electric potential energy = gain in kinetic energy
qV = (0.5) mp v2
(1.6 x 10-19) (19000) = (0.5) (1.67 x 10-27) v2
v = 1.9 x 106 m/s
2.
using right hand rule , direction of magnetic force comes out to be toward left
3.
The proton moves in a semicircle
4.
h = greatest height reached = m v/(qB) = (1.67 x 10-27) (1.9 x 106)/((1.6 x 10-19) (10 x 10-6)) =1983 m
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