Question

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 = $1.6 \times 10^{-19} C)$

• A. $2.0 \times 10^{-24} \; kg $
• B. $1.6 \times 10^{-19} \; kg $
• C. $1.6 \times 10^{-27} \; kg $
• D. $9.1 \times 10^{-31} \; kg $

Answer 1

Answer: (A)

$\frac{mv^{2}}{R} = qvB$
$ 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 $