$A$ charged particle moves through a magnetic field in a direction perpendicular to it. Then the

The radius of the path of an electron moving at a speed of $3.2 \times 10^7 \ m/s$ in a magnetic field of $6 \times 10^{-4} \ T$ perpendicular to it is (mass of electron is $9 \times 10^{-31} \ kg$ and charge of electron is $1.6 \times 10^{-19} \ C$). (in $cm$)

Two charges of same magnitude move in two circles of radii $R_1=R$ and $R_2=2R$ in a region of constant uniform magnetic field $B_0$. The work $W_1$ and $W_2$ done by the magnetic field in the two cases respectively,are such that

An electron is moving in the north direction. It experiences a force in the vertically upward direction. The magnetic field at the position of the electron is in the direction of

In the figure shown,a charge $q$ of mass $m$ moving with a velocity $v$ along the $x$-axis enters a region of uniform magnetic field $B$ directed into the page. If the particle is able to enter the region $x > b$,then the velocity $v$ must be greater than:

$A$ 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: