An electron moves through a uniform magnetic field $\vec{B} = B_0 \hat{i} + 2 B_0 \hat{j} \ T$. At a particular instant of time,the velocity of the electron is $\vec{v} = 3 \hat{i} + 5 \hat{j} \ m/s$. If the magnetic force acting on the electron is $\vec{F} = 5e \hat{k} \ N$,where $e$ is the magnitude of the charge of an electron,then the value of $B_0$ is . . . . . . $T$.

$A$ proton, a deuteron (nucleus of ${ }_1 H^2$) and an $\alpha$-particle with the same kinetic energy enter a region of uniform magnetic field moving at right angles to the field. The ratio of the radii of their circular paths is:

An electron is moving along the positive $X$-axis. You want to apply a magnetic field for a short time so that the electron may reverse its direction and move parallel to the negative $X$-axis. This can be done by applying the magnetic field along

An $\alpha$ particle is moving along a circle of radius $R$ with a constant angular velocity $\omega$. Point $A$ lies in the same plane at a distance $2R$ from the centre. Point $A$ records the magnetic field produced by the $\alpha$ particle. If the minimum time interval between two successive times at which $A$ records zero magnetic field is $t$,find the angular speed $\omega$ in terms of $t$.

An electron moves with speed $2 \times 10^5 \ m/s$ along the positive $x$-direction in the presence of a magnetic field of induction $B = \hat{i} + 4\hat{j} - 3\hat{k} \ T$. The magnitude of the force experienced by the electron in newtons is (Charge on the electron $= 1.6 \times 10^{-19} \ C$)

$A$ charged particle of charge $q$ and mass $m$ gets deflected through an angle $\theta$ upon passing through a square region of side $a$,which contains a uniform magnetic field $B$ normal to its plane. Assuming that the particle entered the square at right angles to one side,what is the speed of the particle?