An electron moves through a uniform magnetic field $\vec{B} = B_0 \hat{i} + 2 B_0 \hat{j} \ T$. At a particular instant of time,the velocity of the electron is $\vec{v} = 3 \hat{i} + 5 \hat{j} \ m/s$. If the magnetic force acting on the electron is $\vec{F} = 5e \hat{k} \ N$,where $e$ is the magnitude of the charge of an electron,then the value of $B_0$ is . . . . . . $T$.

$A$ proton enters a magnetic field of flux density $1.5 \,Wb \,m^{-2}$ with a velocity of $2 \times 10^7 \,ms^{-1}$ at an angle of $30^{\circ}$ with the field. The force on the proton will be

An electron of mass $m$ and charge $e$ is projected normally from the surface of a sphere of radius $a$ with speed $v_0$ into a uniform magnetic field $B$ perpendicular to the plane of the paper. The centre of the sphere is at a distance $b$ from a wall. If the electron strikes the wall symmetrically with respect to the $x$-axis,what should be the magnetic field $B$ such that the charged particle just touches the wall?

Two particles $A$ and $B$ have equal charges but different masses $M_{A}$ and $M_{B}$. After being accelerated through the same potential difference,they enter a region of uniform magnetic field and describe paths of radii $R_{A}$ and $R_{B}$ respectively. Then $M_{A} : M_{B}$ is

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

$A$ charged particle 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: