Maximum kinetic energy of the positive ion in the cyclotron is
$\frac{{{q^2}B{r_0}}}{{2m}}$
$\frac{{q{B^2}{r_o}}}{{2m}}$
$\frac{{{q^2}{B^2}r_0^2}}{{2m}}$
$\frac{{qB{r_0}}}{{2{m^2}}}$
An electron and a positron are released from $(0, 0, 0)$ and $(0, 0, 1.5\, R)$ respectively, in a uniform magnetic field ${\rm{\vec B = }}{{\rm{B}}_0}{\rm{\hat i}}$ , each with an equal momentum of magnitude $P = eBR$. Under what conditions on the direction of momentum will the orbits be non-intersecting circles ?
An electron with energy $0.1\,ke\,V$ moves at right angle to the earth's magnetic field of $1 \times 10^{-4}\,Wbm ^{-2}$. The frequency of revolution of the electron will be. (Take mass of electron $=9.0 \times 10^{-31}\,kg$ )
A particle of specific charge (charge/mass) $\alpha$ starts moving from the origin under the action of an electric field $\vec E = {E_0}\hat i$ and magnetic field $\vec B = {B_0}\hat k$. Its velocity at $(x_0 , y_0 , 0)$ is ($(4\hat i + 3\hat j)$ . The value of $x_0$ is:
charged particle with charge $q$ enters a region of constant, uniform and mutually orthogonal fields $\vec E$ and $\vec B$ with a velocity $\vec v$ perpendicular to both $\vec E$ and $\vec B$ , and comes out without any change in magnitude or direction of $\vec v$ . Then
For a positively charged particle moving in a $x-y$ plane initially along the $x$-axis, there is a sudden change in its path due to the presence of electric and/or magnetic fields beyond $P$. The curved path is shown in the $x-y$ plane and is found to be non-circular. Which one of the following combinations is possible