$A$ negative test charge is moving near a long straight wire carrying a current. The force acting on the test charge is parallel to the direction of the current. The motion of the charge is

$A$ proton is projected with a speed of $2 \times 10^6 \ m/s$ at an angle of $60^{\circ}$ to the $x$-axis. If a uniform magnetic field of $0.104 \ T$ is applied along the $y$-axis,the path of the proton is:

$A$ charged particle with a velocity $2 \times 10^{3} \ ms^{-1}$ passes undeflected through electric and magnetic fields in mutually perpendicular directions. The magnetic field is $1.5 \ T$. The magnitude of the electric field will be:

When an electron placed in a uniform magnetic field is accelerated from rest through a potential difference $V_1$,it experiences a force $F$. If the potential difference is changed to $V_2$,the force experienced by the electron in the same magnetic field is $2F$. Then,the ratio of potential differences $\frac{V_2}{V_1}$ is:

$A$ charged particle moves with some initial velocity along the direction of an external magnetic field $B$. Now,if we apply a uniform electric field $E$ perpendicular to the magnetic field,then the trajectory of the charged particle will be

$A$ particle of charge $1.6 \ \mu C$ and mass $16 \ \mu g$ is present in a strong magnetic field of $6.28 \ T$. The particle is then fired perpendicular to the magnetic field. The time required for the particle to return to its original location for the first time is . . . . . . $s$. $(\pi = 3.14)$