A charged particle moves in a uniform magnetic field. The velocity of the particle at some instant makes an acute angle with the magnetic field. The path of the particle will be

Proton with kinetic energy of $1\;MeV$ moves from south to north. It gets an acceleration of $10^{12}\; \mathrm{m} / \mathrm{s}^{2}$ by an applied magnetic field (west to east). The value of magnetic field :.......$mT$ (Rest mass of proton is $1.6 \times 10^{-27} \;\mathrm{kg}$ )

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 :

If $\alpha $ and $\beta  - $ particles are moving with equal velocity perpendicular to the flux density $B$, then the radii of their paths will be

A positive charge $'q'$ of mass $'m'$ is moving along the $+ x$ axis. We wish to apply a uniform magnetic field $B$ for time $\Delta t$ so that the charge reverses its direction crossing the $y$ axis at a distance $d.$ Then

In the product

$\overrightarrow{\mathrm{F}} =\mathrm{q}(\vec{v} \times \overrightarrow{\mathrm{B}})$

$=\mathrm{q} \vec{v} \times\left(\mathrm{B} \hat{i}+\mathrm{B} \hat{j}+\mathrm{B}_{0} \hat{k}\right)$

For $\mathrm{q}=1$ and $\vec{v}=2 \hat{i}+4 \hat{j}+6 \hat{k}$ and

$\overrightarrow{\mathrm{F}}=4 \hat{i}-20 \hat{j}+12 \hat{k}$

What will be the complete expression for $\vec{B}$ ?