Q. A particle having the same charge as of electron moves in a circular path of radius 0.5 cm under the influence of a magnetic field of 0.5 T. If an electric field of 100 V/m makes it to move in a straight path, then the mass of the particle is (Given charge of electron = $1.6 \times 10^{-19} C)$

Solution:

$\frac{mv^{2}}{R} = qvB$
$ mv =qBR $ ....(i)
Path is straight line
it $ qE = qvB $
$ E = vB $ ....(ii)
From equation (i) & (ii)
$ m = \frac{qB^{2}R}{E} $
$ m = 2.0 \times10^{-24} kg $

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