Maximum kinetic energy of the positive ion in the cyclotron is
$\frac{{{q^2}B{r_0}}}{{2m}}$
$\frac{{q{B^2}{r_o}}}{{2m}}$
$\frac{{{q^2}{B^2}r_0^2}}{{2m}}$
$\frac{{qB{r_0}}}{{2{m^2}}}$
A electron experiences a force $\left( {4.0\,\hat i + 3.0\,\hat j} \right)\times 10^{-13} N$ in a uniform magnetic field when its velocity is $2.5\,\hat k \times \,{10^7} ms^{-1}$. When the velocity is redirected and becomes $\left( {1.5\,\hat i - 2.0\,\hat j} \right) \times {10^7}$, the magnetic force of the electron is zero. The magnetic field $\vec B$ is :
At $t = 0$ a charge $q$ is at the origin and moving in the $y-$ direction with velocity $\overrightarrow v = v\,\hat j .$ The charge moves in a magnetic field that is for $y > 0$ out of page and given by $B_1 \hat z$ and for $y < 0$ into the page and given $-B_2 \hat z .$ The charge's subsequent trajectory is shown in the sketch. From this information, we can deduce that
An $\alpha$-particle (mass $4 amu$ ) and a singly charged sulfur ion (mass $32 amu$ ) are initially at rest. They are accelerated through a potential $V$ and then allowed to pass into a region of uniform magnetic field which is normal to the velocities of the particles. Within this region, the $\alpha$-particle and the sulfur ion move in circular orbits of radii $r_\alpha$ and $r_5$, respectively. The ratio $\left(r_s / r_\alpha\right)$ is. . . . .$(4)$
A particle of charge $q$ and mass $m$ is moving with a velocity $-v \hat{ i }(v \neq 0)$ towards a large screen placed in the $Y - Z$ plane at a distance $d.$ If there is a magnetic field $\overrightarrow{ B }= B _{0} \hat{ k },$ the minimum value of $v$ for which the particle will not hit the screen is
Two ions of masses $4 \,{amu}$ and $16\, amu$ have charges $+2 {e}$ and $+3 {e}$ respectively. These ions pass through the region of constant perpendicular magnetic field. The kinetic energy of both ions is same. Then :