Question

A long straight wire carries a certain current and produces a magnetic field $2 \times 10^{-4} \,Wb\,m ^{2}$ at a perpendicular distance of $5 \,cm$ from the wire. An electron situated at $5 cm$ from the wire moves with a velocity $10^{7} \,m / s$ towards the wire along perpendicular to it. The force experienced by the electron will be (charge on electron $1.6 \times 10^{-19} C$ ) :

• A. $3.2\, NR$
• B. $3.2\times10^{-16}N$
• C. $1.6\times10^{-16}N$
• D. zero

Answer 1

Answer: (B)

The situation is as shown in the figure.

Here, $v=10^{7} m / s , B=2 \times 10^{-4} Wb / m ^{2}$
The magnitude of the force experienced by the
electron is
$F=e v B \sin \theta$
$(\therefore \vec{v} $ and $ \vec{B}$ are perpendicular to each other)
$=e v B \sin 90^{\circ}$
$=1.6 \times 10^{-19} \times 10^{7} \times 2 \times 10^{-4} \times 1$
$=3.2 \times 10^{-16} N$