Question

A long straight wire carries a certain current and produces a magnetic field $2\times {10}^{-4}Wb{m}^{-2}$ at perpendicular distance of 5 cm from the wire. An electron situated at 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 $1.6\times {10}^{-1\circ }C$

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

Answer 1

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

Given magnetic field $\hat{B}=2\times {10}^{-4}Wb{m}^{-2}$

velocity of electron$v={10}^{7}m/s$

charge of electron $q=1.6\times {10}^{-19}C$

We know magnetic force$F=qvB\sin \theta$

$F=1.6\times {10}^{-19}\times {10}^7\times 2\times {10}^{-4}=3.2\times {10}^{-16}$

Here $\theta$=90 as mentioned electrons moving in a perpendicular direction to the wire viz. $\sin 90=1$