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प्रश्न
A particle moves in a circle of diameter 1.0 cm under the action of a magnetic field of 0.40 T. An electric field of 200 V m−1 makes the path straight. Find the charge/mass ratio of the particle.
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उत्तर
Given:
Diameter of the circle = 1.0 cm
Thus, radius of circle, r = = 0.5 × 10−2 m,
Magnetic field, B = 0.40 T
Electric field, E = 200 V m−1.
As per the question, the particle is moving in a circle under the action of a magnetic field. But when an electric field is applied on the particle, it moves in a straight line.
So, we can write:
Fe = Fm
qE = qvB, where q is the charge and v is the velocity of the particle.
⇒ `v = E/B = 200/0.4 = 500 m//s`
As r = `v/(rB)`
=`500/(0.5xx10^-2xx0.4)`
= 2.5 × 105 c /kg
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