## In This post you will get answer of Q.

A very long straight wire carries a current $l$. At the instant when a charge +Q at point P has velocity $overrightarrow{v}$, as shown, the force on the charge is

Solution:

## According to Fleming’s left hand rule direction

of force is along Oy axis which is perpendicular to wire.

$overrightarrow{F} = e(overrightarrow{v} times overrightarrow{B})$.

B due to i is acting inwards i.e. into the paper.

v is along Ox.

$therefore F = Q^+ begin {vmatrix}

widehat{i} & widehat{j} & widehat{k} \

v & 0 & 0 \

0 & 0 & -B end {vmatrix} Rightarrow F = Q^+ [ – widehat{j}(-v B)+ 0 ] $

$therefore overrightarrow{ F} = +Q v B widehat{j}$.

i.e in Oy direction.