A particle having some charge is projected in $x-y$ plane with a speed of $5\ m/s$ in a region having uniform magnetic field along $z-$ axis. Which of the following cannot be the possible value of velocity at any time ?

A particle of mass $m$ and charge $q$ moves with a constant velocity $v$ along the positive $x$ direction. It enters a region containing a uniform magnetic field $B$ directed along the negative $z$ direction, extending from $x = a$ to $x = b$. The minimum value of $v$ required so that the particle can just enter the region $x > b$ is

Two charges of same magnitude move in two circles of radii $R_1=R$ and $R_2=2 R$ 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

A uniform magnetic field $B$ is acting from south to north and is of magnitude $1.5$ $Wb/{m^2}$. If a proton having mass $ = 1.7 \times {10^{ - 27}}\,kg$ and charge $ = 1.6 \times {10^{ - 19}}\,C$ moves in this field vertically downwards with energy $5\, MeV$, then the force acting on it will be

A charged particle carrying charge $1\,\mu C$  is moving with velocity $(2 \hat{ i }+3 \hat{ j }+4 \hat{ k })\, ms ^{-1} .$ If an external magnetic field of $(5 \hat{ i }+3 \hat{ j }-6 \hat{ k }) \times 10^{-3}\, T$ exists in the region where the particle is moving then the force on the particle is $\overline{ F } \times 10^{-9} N$. The vector $\overrightarrow{ F }$ is :

charged particle with charge $q$ enters a region of constant, uniform and mutually orthogonal fields $\vec E$ and $\vec B$ with a velocity $\vec v$ perpendicular to both $\vec E$ and $\vec B$ , and comes out without any change in magnitude or direction of $\vec v$ . Then