$A$ very high magnetic field is applied to a stationary charge. Then the charge experiences

$A$ rectangular region $ABCD$ contains a uniform magnetic field $B_0$ directed perpendicular to the plane of the rectangle. $A$ narrow stream of charged particles moving perpendicularly to the side $AB$ enters this region and is ejected through the adjacent side $BC$ suffering a deflection through $30^{\circ}$. In order to increase this deflection to $60^{\circ}$,the magnetic field has to be

$A$ proton is projected with a speed of $2 \times 10^6 \ m/s$ at an angle of $60^{\circ}$ to the $x$-axis. If a uniform magnetic field of $0.104 \ T$ is applied along the $y$-axis,the path of the proton is:

Under the influence of a uniform magnetic field,a charged particle is moving in a circle of radius $R$ with constant speed $v$. The time period of the motion

An electron $(q = 1.6 \times 10^{-19}\, C)$ is moving at a right angle to a uniform magnetic field of $3.534 \times 10^{-5}\, T$. The time taken by the electron to complete one circular orbit is ...... $\mu s$.

$A$ proton, an electron, and an $\alpha$-particle enter at right angles to a uniform magnetic field with the same velocity. If $R_p$, $R_e$, and $R_{\alpha}$ are the radii of the circular paths of these particles, then: