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 $1.6 \times 10^{-19} \mathrm{C}$ )

• A

  $3.2 \mathrm{~N}$
• Correct answer
  B

  $3.2 \times 10^{-16} \mathrm{~N}$
• C

  $1.6 \times 10^{-16} \mathrm{~N}$
• D

  Zero

Solution

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 $$ \begin{aligned} & =e v \mathrm{~B} \sin 90^{\circ} \\ & =1.6 \times 10^{-19} \times 10^7 \times 2 \times 10^{-4} \times 1 \\ & =3.2 \times 10^{-16} \mathrm{~N} \end{aligned} $$

Hence, the correct answer is option: 'B'.