The electrostatic force $\left(\vec{F}_1\right)$ and magnetic force $\left(\vec{F}_2\right)$ acting on a charge $q$ moving with velocity $v$ can be written :
$\vec{F}_1=q \vec{V} \cdot \vec{E}, \vec{F}_2=q(\vec{B} \cdot \vec{V})$
$\overrightarrow{\mathrm{F}}_1=\mathrm{q} \overrightarrow{\mathrm{B}}, \overrightarrow{\mathrm{F}}_2=\mathrm{q}(\overrightarrow{\mathrm{B}} \times \overrightarrow{\mathrm{V}})$
$\overrightarrow{\mathrm{F}}_1=\mathrm{q} \overrightarrow{\mathrm{E}}, \overrightarrow{\mathrm{F}}_2=\mathrm{q}(\overrightarrow{\mathrm{V}} \times \overrightarrow{\mathrm{B}})$
A long solenoid has $100\,turns/m$ and carries current $i.$ An electron moves with in the solenoid in a circle of radius $2·30\,cm$ perpendicular to the solenoid axis. The speed of the electron is $0·046\,c$ ($c =$ speed of light). Find the current $i$ in the solenoid (approximate).....$A$
A very high magnetic field is applied to a stationary charge. Then the charge experiences
A proton enters a magnetic field of flux density $1.5\,weber/{m^2}$ with a velocity of $2 \times {10^7}\,m/\sec $ at an angle of $30^\circ $ with the field. The force on the proton will be
A particle of mass $m$ and charge $q$, accelerated by a potential difference $V$ enters a region of a uniform transverse magnetic field $B$. If $d$ is the thickness of the region of $B$, the angle $\theta$ through which the particle deviates from the initial direction on leaving the region is given by
A charged particle is released from rest in a region of steady uniform electric and magnetic fields which are parallel to each other the particle will move in a