Inside a region of non uniform magnetic field given by the formula $\overrightarrow{B}=-B_0 x{k}^{^}$, a particle of mass $M$ and charge$q$ is projected horizontally as shown. It is given that $q{B}_0/M =1$. At any instant the velocity vector of the particle is found to $\left(\raisebox{1ex}{$v$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\right)i^{^}+a{j}^{^}+b{k}^{^}$ If the initial value of the velocity is $v \raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s$}\right.$, then

As the magnetic field is acting along $z-axis$, and particle initial velocity is along $x-axis$, so particle will remain in $xy-plane$. Also its speed must be constant. Thus

$$\begin{gathered} b=0 and v = \sqrt{{v^2}_x+{v^2}_y} \\ or v^2={\left(\frac{v}{2}\right)}^2+a^2 \\ \therefore a =\frac{\sqrt{3}}{2}v \\ Now, qvB= M\left(v\frac{dv}{dx}\right) \\ or qv\times \left(-B_{0}x\right) = Mv\frac{dv}{dx} \\ or -{\int }_v^{0}dv=\frac{q{B}_0}{M}{\int }_0^{x}xdx \\ or v = \left(\frac{q{B}_0}{M}\right)\frac{x^2}{2} \\ or v = 1\times \frac{x^2}{2} \\ \therefore x = \sqrt{2v} \end{gathered}$$