A particle of charge $q$, mass $m$ enters in a region of magnetic field $B$ with velocity $V_0 \widehat i$. Find the value of $d$ if the particle emerges from the region of magnetic field at an angle $30^o$ to its ititial velocity:-

A particle is moving in a uniform magnetic field, then

A magnetic field $\overrightarrow{\mathrm{B}}=\mathrm{B}_0 \hat{\mathrm{j}}$ exists in the region $\mathrm{a} < \mathrm{x} < 2 \mathrm{a}$ and $\vec{B}=-B_0 \hat{j}$, in the region $2 \mathrm{a} < \mathrm{x} < 3 \mathrm{a}$, where $\mathrm{B}_0$ is a positive constant. $\mathrm{A}$ positive point charge moving with a velocity $\overrightarrow{\mathrm{v}}=\mathrm{v}_0 \hat{\dot{i}}$, where $v_0$ is a positive constant, enters the magnetic field at $x=a$. The trajectory of the charge in this region can be like,

Given below are two statements: One is labelled as Assertion $(A)$ and the other is labelled as Reason $(R).$

Assertion $(A)$ : In an uniform magnetic field, speed and energy remains the same for a moving charged particle.

Reason $(R)$ : Moving charged particle experiences magnetic force perpendicular to its direction of motion.

In a mass spectrometer used for measuring the masses of ions, the ions are initially accelerated by an electric potential $V$ and then made to describe semicircular paths of radius $R$ using a magnetic field $B$. If $V$ and $B$ are kept constant, the ratio $\left( {\frac{{{\text{charge on the ion}}}}{{{\text{mass of the ion}}}}} \right)$ will be proportional to

A proton (mass $ = 1.67 \times {10^{ - 27}}\,kg$ and charge $ = 1.6 \times {10^{ - 19}}\,C)$ enters perpendicular to a magnetic field of intensity $2$ $weber/{m^2}$ with a velocity $3.4 \times {10^7}\,m/\sec $. The acceleration of the proton should be