Q. 484.5( 2 Votes )

A proton goes undeflected in a crossed electric and magnetic field (the fields are perpendicular to each other) at a speed of 2.0 × 106 ms–1. The velocity is perpendicular to both the fields. When the electric field is switched off, the proton moves along a circle of radius 4.0 cm. Find the magnitudes of the electric and the magnetic fields. Take the mass of the proton = 1.6 × 10–27 kg.

Answer :

Given-
Mass of the proton, m = 1.6 × 10−27 kg



Speed of the proton inside the crossed electric and magnetic field, v = 2.0 × 105 ms−1



Given from question we can confer that , the proton is not deflected under the combined action of the electric and magnetic fields.


Thus, the forces applied by both the fields are equal and opposite to each other


Magnetic force, we know, Lorentz force F is given by -



where,


q = charge on an electron


v = velocity of the electron


B=magnetic field


θ= angle between B and v


Also,


We, know the Columb’s force–



where


E = applied electric field


e=charge



Now,






But when the electric field is switched off, the proton moves follows circular path due to the presence of force of the magnetic field.



The radius of the circular path described by a particle in a magnetic field r,



where,


m is the mass of a proton


v= velocity of the particle


B = magnetic force


q= charge on the particle = C





Substituting the value of B in equation (1), we get





Hence magnitudes of the electric and the magnetic fields are



respectively


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