$A$ proton and an $\alpha$-particle enter a uniform magnetic field perpendicularly with the same speed. If a proton takes $25 \ \mu s$ to make $5$ revolutions,then the periodic time for the $\alpha$-particle would be ........ $\mu s$.

An electron moving with a uniform velocity along the positive $x$-direction enters a magnetic field directed along the positive $y$-direction. The force on the electron is directed along

$A$ proton and an $\alpha$-particle are simultaneously projected in opposite directions into a region of uniform magnetic field of $2 \text{ mT}$ perpendicular to the direction of the field. After some time,it is found that the velocity of the proton has changed in direction by $90^{\circ}$. Then,at this time,the angle between the velocity vectors of the proton and the $\alpha$-particle is (in $^{\circ}$)

$A$ particle moving in a magnetic field increases its velocity,then its radius of the circle

Two charges of same magnitude move in two circles of radii $R_1=R$ and $R_2=2R$ in a region of constant uniform magnetic field $B_0$. The work $W_1$ and $W_2$ done by the magnetic field in the two cases respectively,are such that

An $\alpha$-particle travels in a circular path of radius $0.45\,m$ in a magnetic field $B = 1.2\,Wb/m^2$ with a speed of $2.6 \times 10^7\,m/s$. The period of revolution of the $\alpha$-particle is: