$A$ velocity selector is to be constructed to select ions with a velocity of $6 \,km \,s^{-1}$. If the electric field used is $400 \,V \,m^{-1}$, then the magnetic field to be used is

$A$ charged particle of charge '$q$' is accelerated by a potential difference '$V$' and enters a region of uniform magnetic field '$B$' at right angles to the direction of the field. The charged particle completes a semicircle of radius '$r$' inside the magnetic field. The mass of the charged particle is

Describe the features of the magnetic force acting on a charged particle moving inside a magnetic field.

An electron having kinetic energy $T$ is moving in a circular orbit of radius $R$ perpendicular to a uniform magnetic induction $\vec{B}$. If kinetic energy is doubled and magnetic induction is tripled,the radius will become

$A$ nucleus at rest undergoes $\alpha -$ decay according to the equation ${}_{92}^{226}X \to Y + \alpha$. At $t=0$,the emitted $\alpha -$ particle enters a region of space where a uniform magnetic field $\vec{B} = B_0 \hat{i}$ and an electric field $\vec{E} = E_0 \hat{i}$ exist. The $\alpha -$ particle enters the region with velocity $\vec{v} = v_0 \hat{j}$ at $x = 0$. After a time $t = \left( \sqrt{3} \times 10^7 \frac{m_{\alpha}}{q_{\alpha} E_0} \right)$,the particle is found to have twice its initial speed. The initial speed $v_0$ of the $\alpha -$ particle is:

$A$ proton (mass $m$,charge $q$) accelerated by a potential difference $V$ flies through a uniform transverse magnetic field $B$. The field occupies a region of space of width $d$. If $\alpha$ is the angle of deviation of the proton from the initial direction of motion (see figure),the value of $\sin \alpha$ will be: