Question
A particle of mass m and charge q enters a uniform Magnetic field B (perpendicular to the paper in wards) at P with velocity V at an angle α and leaves the field at Q at an angle β as shown. Explain how these options are correct?
1) α must be equal to β,
2) Length PQ=(2mv*Sin α)/qB,
3) Particle remains in the field for time T=2mα/(qB).
Answer
1)
Since velocity is perpendicular to the direction of magnetic field, the charge experiences force in a direction perpendicular to both. As the force is perpendicular to the velocity, it does no work and results in uniform circular motion of the charge. If the angles α and β are the same, it means that the velcocity of the charge at both these points P and Q is tangential to the circular arc between P and Q inside the magnetic field which is the path of motion of the charge.
2)
The force acting on the charge
= qvB
which equals the centripetal force, making it move on the circular path of radius R given by
qvB = mv^2/R
=> R = mv/qB ... ( 1 )
A perpendicular from the center of the circular path on the chord PQ bisects it at M
PQ
= 2 PM
= 2 Rsinα
Plugging the value of R from ( 1 ) above,
PQ = (2mvsinα)/qB
3)
Arc length PQ
= 2αR
= 2α * mv/(qb)
Time taken to travel this distance with constant velocity, v is
T = distance /velocity
= [2αmv/(qB)]/v
= 2mα/(qB).
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