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Karnataka Board PUCPUC Science Class 11

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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Question

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.

Sum
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Solution

Given:
Diameter of the circle = 1.0 cm
Thus, radius of circle, = = 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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Force on a Moving Charge in Uniform Magnetic and Electric Fields
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Chapter 12: Magnetic Field - Exercises [Page 234]

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HC Verma Concepts of Physics Vol. 2 [English] Class 11 and 12
Chapter 12 Magnetic Field
Exercises | Q 47 | Page 234
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