A charged particle of charge $q$ and mass $m$, gets deflected through an angle $\theta$ upon passing through a square region of side $a$, which contains a uniform magnetic field $B$ normal to its plane. Assuming that the particle entered the square at right angles to one side, what is the speed of the particle?
$\frac{q B}{m} a \cot \theta$
$\frac{q B}{m} a \tan \theta$
$\frac{q B}{m} a \cot ^2 \theta$
$\frac{q B}{m} \alpha \tan ^2 \theta$
A particle of charge $q$ and mass $m$ is moving along the $x$ -axis with a velocity $v$ and enters a region of electric field $E$ and magnetic field $B$ as shown in figure below for which figure the net force on the charge may be zero
Consider a thin metallic sheet perpendicular to the plane of the paper moving with speed $'v'$ in a uniform magnetic field $B$ going into the plane of the paper (See figure). If charge densities ${\sigma _1}$ and ${\sigma _2}$ are induced on the left and right surfaces, respectively, of the sheet then (ignore fringe effects)
Show that a force that does no work must be a velocity dependent force.
In a chamber, a uniform magnetic field of $6.5 \;G \left(1 \;G =10^{-4} \;T \right)$ is maintained. An electron is shot into the field with a speed of $4.8 \times 10^{6} \;m s ^{-1}$ normal to the field.the radius of the circular orbit of the electron is $4.2 \;cm$. obtain the frequency of revolution of the electron in its circular orbit. Does the answer depend on the speed of the electron? Explain.
$\left(e=1.5 \times 10^{-19} \;C , m_{e}=9.1 \times 10^{-31}\; kg \right)$
A charged particle is moving in a uniform magnetic field in a circular path. Radius of circular path is $R$. When energy of particle is doubled, then new radius will be