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 charged particle is released from rest in a region of uniform electric and magnetic fields which are parallel to each other. The particle will move on a
An electron emitted by a heated cathode and accelerated through a potential difference of $ 2.0 \;kV$, enters a region with uniform magnetic field of $0.15\; T$. Determine the trajectory of the electron if the field
$(a)$ is transverse to its initial velocity,
$(b)$ makes an angle of $30^o$ with the initial velocity
A particle with charge $+Q$ and mass m enters a magnetic field of magnitude $B,$ existing only to the right of the boundary $YZ$. The direction of the motion of the $m$ particle is perpendicular to the direction of $B.$ Let $T = 2\pi\frac{m}{{QB}}$ . The time spent by the particle in the field will be
A charged particle is moving in a circular orbit of radius $6\, cm$ with a uniform speed of $3 \times 10^6\, m/s$ under the action of a uniform magnetic field $2 \times 10^{-4}\, Wb/m^2$ which is at right angles to the plane of the orbit. The charge to mass ratio of the particle is
A proton is projected with a velocity $10^7\, m/s$, at right angles to a uniform magnetic field of induction $100\, mT$. The time (in second) taken by the proton to traverse $90^o$ arc is $(m_p = 1.65\times10^{-27}\, kg$ and $q_p = 1.6\times10^{-19}\, C)$