or $\mathrm{ma}_{0}=\mathrm{eE}$ or, $\mathrm{E}=\mathrm{ma} / \mathrm{e}$ towards west
The acceleration changes from a to $3 \mathrm{a}_{0}$
Hence net acceleration produced by magnetic fleld vector $\mathrm{B}$ is $2 \mathrm{a}_{0}$
Force due to magnetic field
$=\overrightarrow{\mathrm{F}_{\mathrm{B}}}=\mathrm{m} \times 2 \mathrm{a}_{0}=\mathrm{e} \times \mathrm{V}_{0} \times \mathrm{B}$
$\Rightarrow \mathrm{B}=\frac{2 \mathrm{ma}_{0}}{\mathrm{eV}_{0}} \quad$ downwards


A charge particle of $3 \pi$ coulomb is passing through the point $P$ with velocity
$\overrightarrow{ v }=(2 \hat{ i }+3 \hat{ j }) \,m / s$; where $\hat{i}$ and $\hat{j} \quad$ represents unit vector along $x$ and $y$ axis respectively.
The force acting on the charge particle is $4 \pi \times 10^{-5}(-x \hat{i}+2 \hat{j}) \,N$. The value of $x$ is