The acceleration of an electron at a moment in a magnetic field $\vec{B} = 2\hat{i} + 3\hat{j} + 4\hat{k}$ is $\vec{a} = x\hat{i} - 2\hat{j} + \hat{k}$. The value of $x$ is

Two ions having masses in the ratio $1: 1$ and charges $1: 2$ are projected into a uniform magnetic field perpendicular to the field with speeds in the ratio $2: 3$. The ratio of the radii of circular paths along which the two particles move is :

$A$ proton of energy $200\, MeV$ enters a magnetic field of $5\, T$. If the direction of the field is from south to north and the motion is upward,the force acting on it will be:

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:

An electron (mass = $9.0 \times 10^{-31} \ kg$ and charge = $1.6 \times 10^{-19} \ C$) is moving in a circular orbit in a magnetic field of $1.0 \times 10^{-4} \ Wb/m^2$. Its period of revolution is:

$A$ particle of mass $m = 1.67 \times 10^{-27} \, kg$ and charge $q = 1.6 \times 10^{-19} \, C$ enters a region of uniform magnetic field of strength $B = 1 \, T$ as shown in the figure. The magnetic field is directed into the plane of the paper. The particle enters at point $C$ with an angle of $45^{\circ}$ with the boundary. Find the time spent by the particle in the magnetic field region in $ns$.