Question

An electron is moving along x-axis in x-y plane with velocity of ${10}^{2}m/s$. A magnetic field of magnitude $10T$ is present in the region along z-axis. If the magnitude of force on the particle is $1.6m\times {10}^{-16}N$, find $m$

Answer 1

• Magnetic force is given as $F=qvBsin\theta$ it is given that angle between field $B$ and velocity $v$ is ${90}^0$ and $q=1.6\times {10}^{-19}C$ so we get the value of force as $F=1.6\times {10}^{-16}N$ so $m=1$ is the required answer.