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A long straight wire carries a certain current and produces a magnetic field $2 \times 10^{-4} \frac{\text { Weber }}{\mathrm{m}^2}$ at a perpendicular distance of $5 \mathrm{~cm}$ from the wire. An electron situated at $5 \mathrm{~cm}$ from the wire moves with a velocity $10^7 \mathrm{~m} / \mathrm{s}$ towards the wire along perpendicular to it. The force experienced by the electron will be (change on electron $1.6 \times 10^{-19} \mathrm{C}$ )
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Verified Answer
The correct answer is:
$3.2 \times 10^{-16} \mathrm{~N}$
From the question
Velocity of electron $=10^7 \mathrm{~m} / \mathrm{s}$
Magnetic field $B=2 \times 10^{-4} \mathrm{~Wb} / \mathrm{m}^2$
Then magnitude of force experienced by the electron i.e.
$$
F=e v \mathrm{~B} \sin \theta
$$
$\{\because V$ and $B$ are prependicular to each other
$$
\begin{aligned}
& =e v \mathrm{~B} \sin 90^{\circ} \\
& =1.6 \times 10^{-19} \times 10^7 \times 2 \times 10^{-4} \times 1 \\
& =3.2 \times 10^{-16} \mathrm{~N}
\end{aligned}
$$
Velocity of electron $=10^7 \mathrm{~m} / \mathrm{s}$
Magnetic field $B=2 \times 10^{-4} \mathrm{~Wb} / \mathrm{m}^2$
Then magnitude of force experienced by the electron i.e.
$$
F=e v \mathrm{~B} \sin \theta
$$
$\{\because V$ and $B$ are prependicular to each other
$$
\begin{aligned}
& =e v \mathrm{~B} \sin 90^{\circ} \\
& =1.6 \times 10^{-19} \times 10^7 \times 2 \times 10^{-4} \times 1 \\
& =3.2 \times 10^{-16} \mathrm{~N}
\end{aligned}
$$
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