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. A particle say 'c' having charge equal to that of an electron is situated at $5$ $cm$ from the wire moves with a velocity of ${10}^{7}m/s$ towards the wire perpendicularly. The magnitude of force experienced by the particle will be (charge of an electron =$-1.6\times {10}^{-19}C$ )

• A. $3.2N$
• B. $3.2\times {10}^{-16}N$
• C. $1.6\times {10}^{-16}N$
• D. $0$ $N$

Answer 1

The correct option is B $3.2\times {10}^{-16}N$

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
$F=evB\sin \left(\theta \right)$
($\because \overrightarrow{v}and\overrightarrow{B}$ are perpendicular to each other)
$=evB\sin \left({90}^0\right)=-1.6\times {10}^{-19}\times {10}^7\times 2\times {10}^{-4}\times 1=-3.2\times {10}^{-16}N$
Here, negative sign denotes the direction of force.