If an electron and a proton having same momenta enter perpendicular to a magnetic field, then

A uniform beam of positively charged particles is moving with a constant velocity parallel to another beam of negatively charged particles moving with the same velocity in opposite direction separated by a distance $d.$ The variation of magnetic field $B$ along a perpendicular line draw between the two beams is best represented by

A proton and an $\alpha -$ particle (with their masses in the ratio of $1 : 4$ and charges in the ratio of $1:2$ are accelerated from rest through a potential difference $V$. If a uniform magnetic field $(B)$ is set up perpendicular to their velocities, the ratio of the radii $r_p : r_{\alpha }$ of the circular paths described by them will be

A particle of specific charge (charge/mass) $\alpha$ starts moving from the origin under the action of an electric field $\vec E = {E_0}\hat i$ and magnetic field $\vec B = {B_0}\hat k$. Its velocity at $(x_0 , y_0 , 0)$ is ($(4\hat i + 3\hat j)$ . The value of $x_0$ is: 

A current carrying long solenoid is placed on the ground with its axis vertical. A proton is falling along the axis of the solenoid with a velocity $v$. When the proton enters into the solenoid, it will

An electron is projected with velocity $v_0$ in a uniform electric field $E$ perpendicular to the field. Again it is projetced with velocity $v_0$ perpendicular to a uniform magnetic field $B/$ If $r_1$ is initial radius of curvature just after entering in the electric field and $r_2$ is initial radius of curvature just after entering in magnetic field then the ratio $r_1:r_2$ is equal to