$A$ horizontal conductor is oriented north-south and carries some current. $A$ positively charged particle located vertically above it and having a velocity directed northward experiences an upward force. What is the direction of the force if this charged particle were located to the east of the conductor and had a velocity directed towards the conductor?

$A$ proton of velocity $v = (3 \hat{i} + 2 \hat{j}) \ m/s$ enters a magnetic field of induction $B = (2 \hat{j} + 3 \hat{k}) \ T$. The acceleration produced in the proton in $m/s^2$ is (Specific charge of proton $= 0.96 \times 10^8 \ C/kg$)

An electron having mass $m$ and kinetic energy $K$ enters a uniform magnetic field $B$ perpendicularly. Its frequency will be:

$A$ magnetic field set up using Helmholtz coils is uniform in a small region and has a magnitude of $0.75 \;T$. In the same region,a uniform electrostatic field is maintained in a direction normal to the common axis of the coils. $A$ narrow beam of (single species) charged particles all accelerated through $15 \;kV$ enters this region in a direction perpendicular to both the axis of the coils and the electrostatic field. If the beam remains undeflected when the electrostatic field is $9.0 \times 10^{5} \;V \,m^{-1}$,make a simple guess as to what the beam contains. Why is the answer not unique?

$A$ particle of mass $m$ and charge $q$ is incident on the $XZ$ plane with velocity $v$ in a direction making an angle $\theta$ with a uniform magnetic field applied along the $X$-axis. The nature of motion performed by the particle is . . . . . . .

Two particles $A$ and $B$ of masses $m_A$ and $m_B$ respectively,having the same charge,are moving in a plane. $A$ uniform magnetic field exists perpendicular to this plane. The speeds of the particles are $v_A$ and $v_B$ respectively,and the trajectories are as shown in the figure. Then: