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When a proton is released from rest in a room, it starts with an initial acceleration $a_0$ towards west. When it is projected towards north with a speed $v_0$ it moves with an initial acceleration $3a_0$ toward west. The electric and magnetic fields in the room are
$\frac{{m{a_0}}}{e}$ west , $\frac{{m{a_0}}}{{e{V_0}}}$ up
$\;\frac{{m{a_0}}}{e}$ west , $\frac{{2m{a_0}}}{{e{V_0}}}$ down
$\frac{{m{a_0}}}{e}$ east , $\frac{{3m{a_0}}}{{e{V_0}}}$ up
$\frac{{m{a_0}}}{e}$ east, $\frac{{3m{a_0}}}{{e{V_0}}}$ down
Solution
$\mathrm{q}_{\mathrm{p}}=\mathrm{e}, \mathrm{mp}=\mathrm{m}, \mathrm{F}=\mathrm{q}_{\mathrm{P}} \times \mathrm{E}$
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
