Show that a force that does no work must be a velocity dependent force.
As work done by force is zero,
$\therefore d \mathrm{~W}=\overrightarrow{\mathrm{F}} \cdot \overrightarrow{d l}=0$
$\therefore \overrightarrow{\mathrm{F}} \cdot \overrightarrow{d l} \cdot d t$
$\therefore \overrightarrow{\mathrm{F}} \cdot(\vec{v} \cdot \overrightarrow{d l})=0$
$\therefore \overrightarrow{\mathrm{F}} \cdot \vec{v}=0, d l \neq 0$
$\therefore$ F $v \cos \theta=0$
If $v$ changes direction then to make $\theta=90, \mathrm{~F}$ must charne angle according to $v .$ So, $\mathrm{F}$ is dependent on $v$ to make work done zero.
A proton carrying $1\, Me V$ kinetic energy is moving in a circular path of radius $R$ in uniform magnetic field. What should be the energy of an $\alpha -$ particle to describe a circle of same radius in the same field ?........$MeV$
An electron moves straight inside a charged parallel plate capacitor of uniform charge density. The space between the plates is filled with uniform magnetic field of intensity $B ,$ as shown in the figure, Neglecting effect of gravity, the time of straight line motion of the electron in the capacitor is
A particle of charge $q$ and mass $m$ starts moving from the origin under the action of an electric field $\vec E = {E_0}\hat i$ and $\vec B = {B_0}\hat i$ with velocity ${\rm{\vec v}} = {{\rm{v}}_0}\hat j$. The speed of the particle will become $2v_0$ after a time
An electron moves with a speed of $2 \times 10^5\, m/s$ along the $+ x$ direction in a magnetic field $\vec B = \left( {\hat i - 4\hat j - 3\hat k} \right)\,tesla$. The magnitude of the force (in newton) experienced by the electron is (the charge on electron $= 1.6 \times 10^{-19}\, C$)
Under the influence of a uniform magnetic field a charged particle is moving in a circle of radius $R$ with constant speed $v$. The time period of the motion