Which law is useful to determine relation between current and magnetic fields due to it.

Two particles $\mathrm{X}$ and $\mathrm{Y}$ having equal charges are being accelerated through the same potential difference. Thereafter they enter normally in a region of uniform magnetic field and describes circular paths of radii $R_1$ and $R_2$ respectively. The mass ratio of $\mathrm{X}$ and $\mathrm{Y}$ is :

The figure shows a region of length $'l'$ with a uniform magnetic field of $0.3\, T$ in it and a proton entering the region with velocity $4 \times 10^{5}\, ms ^{-1}$ making an angle $60^{\circ}$ with the field. If the proton completes $10$ revolution by the time it cross the region shown, $l$ is close to....... $m$

(mass of proton $=1.67 \times 10^{-27} \,kg ,$ charge of the proton $\left.=1.6 \times 10^{-19}\, C \right)$

An electron moves through a uniform magnetic field $\vec{B}=B_0 \hat{i}+2 B_0 \hat{j} T$. At a particular instant of time, the velocity of electron is $\overrightarrow{\mathrm{u}}=3 \hat{i}+5 \hat{j} \mathrm{~m} / \mathrm{s}$. If the magnetic force acting on electron is $\vec{F}=5 e\hat kN$, where $e$ is the charge of electron, then the value of $\mathrm{B}_0$ is ____$\mathrm{T}$.

A charged particle enters a uniform magnetic field perpendicular to it. The magnetic field

A uniform magnetic field $B$ is acting from south to north and is of magnitude $1.5$ $Wb/{m^2}$. If a proton having mass $ = 1.7 \times {10^{ - 27}}\,kg$ and charge $ = 1.6 \times {10^{ - 19}}\,C$ moves in this field vertically downwards with energy $5\, MeV$, then the force acting on it will be