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.
Electron moves at right angles to a magnetic field of $1.5 \times 10^{-2}\,tesla$ with speed of $6 \times 10^7\,m/s$. If the specific charge of the electron is $1.7 \times 10^{11}\,C/kg$. The radius of circular path will be......$cm$
An electron enters the space between the plates of a charged capacitor as shown. The charge density on the plate is $\sigma $. Electric intensity in the space between the plates is $E$. A uniform magnetic field $B$ also exists in that space perpendicular to the direction of $E$. The electron moves perpendicular to both $\vec E$ and $\vec B$ without any change in direction. The time taken by the electron to travel a distance $\ell $ is the space is
The magnetic field is uniform for $y>0$ and points into the plane. The magnetic field is uniform and points out of the plane for $y<0$. A proton denoted by filled circle leaves $y=0$ in the $-y$-direction with some speed as shown below.Which of the following best denotes the trajectory of the proton?
A particle with charge to mass ratio, $\frac{q}{m} = \alpha $ is shot with a speed $v$ towards a wall at a distance $d$ perpendicular to the wall. The minimum value of $\vec B$ that exist in this region perpendicular to the projection of velocity for the particle not to hit the wall is
When a charged particle moving with velocity $\vec V$ is subjected to a magnetic field of induction $\vec B$ , the force on it is non-zero. This implies the