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A very long straight wire carries a current I. At the instant when a charge $+Q$ at point $\mathrm{P}$ has velocity $\bar{v}$ as shown the force on the charge is:
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Verified Answer
The correct answer is:
along y
According to Fleming's left hand rule direction of force is along $y$ axis when is perpendicular to wire:
$$
\vec{F}=Q(\vec{V} \times \vec{B})
$$
$\mathrm{B}$ due to $\mathrm{I}$ is acting inwards is into the paper, $v$ is along or:
$$
\begin{array}{ll}
\therefore & F=a+\left|\begin{array}{ccc}
\hat{i} & \hat{h} & \hat{k} \\
v & 0 & 0 \\
0 & 0 & -\mathrm{B}
\end{array}\right| \\
\Rightarrow & F=Q+[-j(-v B)+0] \\
\therefore & F=Q v B \hat{j}
\end{array}
$$
in $y$ direction.
$$
\vec{F}=Q(\vec{V} \times \vec{B})
$$
$\mathrm{B}$ due to $\mathrm{I}$ is acting inwards is into the paper, $v$ is along or:
$$
\begin{array}{ll}
\therefore & F=a+\left|\begin{array}{ccc}
\hat{i} & \hat{h} & \hat{k} \\
v & 0 & 0 \\
0 & 0 & -\mathrm{B}
\end{array}\right| \\
\Rightarrow & F=Q+[-j(-v B)+0] \\
\therefore & F=Q v B \hat{j}
\end{array}
$$
in $y$ direction.
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