The magnetic force acting on a charged particle of charge $-2\, \mu C$ in a magnetic field of $2\, T$ acting in the $y$-direction,when the particle velocity is $(2\hat{i} + 3\hat{j}) \times 10^6\, m/s$,is:

$A$ stream of electrons is projected horizontally to the right. $A$ straight conductor carrying a current is supported parallel to the electron stream and above it. If the current in the conductor is from left to right,then what will be the effect on the electron stream?

$A$ proton beam enters a magnetic field of $ 10^{-4} \,Wb m^{-2} $ normally. If the specific charge of the proton is $ 10^{11} \,C kg^{-1} $ and its velocity is $ 10^{9} \,ms^{-1} $, then the radius of the circle described will be (in $\,m$)

$A$ proton,an electron,and a Helium nucleus have the same kinetic energy. They are moving in circular orbits in a plane due to a magnetic field perpendicular to the plane. Let $r_p, r_e$,and $r_{He}$ be their respective radii. Then:

$A$ charged particle moves in a magnetic field $\vec B = 10\,\hat i$ with initial velocity $\vec u = 5\hat i + 4\hat j$. The path of the particle will be:

$A$ proton is moving undeflected in a region of crossed electric and magnetic fields at a constant speed of $2 \times 10^5 \text{ ms}^{-1}$. When the electric field is switched off,the proton moves along a circular path of radius $2 \text{ cm}$. The magnitude of the electric field is $x \times 10^4 \text{ N/C}$. The value of $x$ is . . . . . . . (Take the mass of the proton $= 1.6 \times 10^{-27} \text{ kg}$ and charge $e = 1.6 \times 10^{-19} \text{ C}$)