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 :

An electron moving with a speed $u$ along the positive $x-$axis at $y = 0$ enters a region of uniform magnetic field $\overrightarrow B = – {B_0}\hat k$ which exists to the right of $y$-axis. The electron exits from the region after some time with the speed $v$ at co-ordinate $y$, then

Under the influence of a uniform magnetic field a charged particle is moving in a circle of radius $R$ with constant speed $v$. The time period of the motion

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

In the product

$\overrightarrow{\mathrm{F}} =\mathrm{q}(\vec{v} \times \overrightarrow{\mathrm{B}})$

$=\mathrm{q} \vec{v} \times\left(\mathrm{B} \hat{i}+\mathrm{B} \hat{j}+\mathrm{B}_{0} \hat{k}\right)$

For $\mathrm{q}=1$ and $\vec{v}=2 \hat{i}+4 \hat{j}+6 \hat{k}$ and

$\overrightarrow{\mathrm{F}}=4 \hat{i}-20 \hat{j}+12 \hat{k}$

What will be the complete expression for $\vec{B}$ ?