$A$ 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:

The charged particle moving in a uniform magnetic field of $(3\hat{i} + 2\hat{j}) \ \text{T}$ has an acceleration $(4\hat{i} - \frac{x}{2}\hat{j}) \ \text{m/s}^2$. The value of $x$ is . . . . . . .

$A$ proton is moving with a uniform velocity of $10^{6} \ m/s$ along the $Y$-axis, under the joint action of a magnetic field along the $Z$-axis and an electric field of magnitude $2 \times 10^{4} \ V/m$ along the negative $X$-axis. If the electric field is switched off, the proton starts moving in a circle. The radius of the circle is nearly: (Given: $\frac{e}{m}$ ratio for proton $= 10^{8} \ C/kg$) (in $m$)

If $\alpha$ and $\beta$ particles are moving with equal velocity perpendicular to the magnetic flux density $B$,then the radii of their paths will be

Work done by a magnetic field of $1\,T$ on a particle of charge $1\,C$ moving with a speed of $5\,m/s$ in one second is........$J$.

An electron is travelling horizontally towards the east. $A$ magnetic field in the vertically downward direction exerts a force on the electron along: