$A$ proton carrying $1\, MeV$ kinetic energy is moving in a circular path of radius $R$ in a uniform magnetic field. What should be the energy of an $\alpha$-particle to describe a circle of the same radius in the same field? ........$MeV$

$A$ particle of mass $m = 1.67 \times 10^{-27} \, kg$ and charge $q = 1.6 \times 10^{-19} \, C$ enters a region of uniform magnetic field of strength $B = 1 \, T$ along the direction shown in the figure. The angle of incidence is $45^{\circ}$ and the angle of emergence is also $45^{\circ}$. The time spent by the particle in the magnetic field is $...... \, ns$.

$A$ particle of charge $1.0 \times 10^{-16} \ C$ moves through a uniform magnetic field $B=B_0(\hat{i}+4 \hat{j}) \ T$. The particle velocity at some instant is $v=(2 \hat{i}+4 \hat{j}) \ ms^{-1}$ and the magnetic force acting on it is $3 \times 10^{-16} \hat{k} \ N$. The magnitude of $B_0$ is (in $T$)

An electron enters a magnetic field whose direction is perpendicular to the velocity of the electron. Then

$A$ capacitor with surface charge density $\sigma$ has a magnetic field $\overrightarrow{B}$ between its two plates. If an electron passes through without any deflection,how long will it take to exit the plates of length $l$?

$A$ 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: