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 $1.6 \times 10^{-27} \;\mathrm{kg}$ )

The dimension of the magnetic field intensity $B$ is

A particle is moving in a uniform magnetic field, then

A small block of mass $m$, having charge $q$ is placed on frictionless inclined plane making an angle  $\theta$ with the horizontal. There exists a uniform magnetic field $B$ parallel to the inclined plane but  perpendicular to the length of spring. If $m$ is slightly pulled on the inclined in downward direction and released, the time period of oscillation will be (assume that the block does not leave contact with the plane)

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

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 :