$A$ charged particle is moving in a uniform magnetic field $(2 \hat{i} + 3 \hat{j}) \text{ T}$. If it has an acceleration of $(\alpha \hat{i} - 4 \hat{j}) \text{ m/s}^2$,then the value of $\alpha$ will be:

$A$ charge $Q$ moves parallel to a very long straight wire carrying a current $I$ as shown. The force on the charge is

$A$ proton of energy $8\, eV$ is moving in a circular path in a uniform magnetic field. The energy of an alpha particle moving in the same magnetic field and along the same path will be.....$eV$

$A$ particle of mass $m$ and charge $q$ enters a magnetic field $B$ perpendicularly with a velocity $v$. The radius of the circular path described by it will be:

$A$ proton (mass $m$,charge $q$) accelerated by a potential difference $V$ flies through a uniform transverse magnetic field $B$. The field occupies a region of space of width $d$. If $\alpha$ is the angle of deviation of the proton from the initial direction of motion (see figure),the value of $\sin \alpha$ will be:

$A$ particle with charge-to-mass ratio $\frac{q}{m} = \alpha$ is shot with a speed $v$ towards a wall at a distance $d$,moving perpendicular to the wall. The minimum value of the magnetic field $\vec{B}$ that exists in this region,perpendicular to the velocity of the particle,such that the particle does not hit the wall is: