Q. A straight conductor carries a current of $5A$. An electron travelling with a speed f $5 \times 10^{6}ms^{- 1}$ parallel to the wire at a distance of $0.1m$ from the conductor, experiences a force of
A) $8 \times 10^{- 20}N$
B) $3.2 \times 10^{- 19}N$
C) $8 \times 10^{- 18}N$
D) $1.6 \times 10^{- 19}N$
Explanation
(c) Magnetic field produced by wire is perpendicular to the motion of electron and it is given by
$B = \frac{\mu_{0}}{4\pi}.\frac{2i}{a} = 10^{- 7} \times \frac{2 \times 5}{0.1} = 10^{- 5}Wb/m^{2}$
Hence force on electron
$F = qvB = \left( 1.6 \times 10^{- 19} \right) \times 5 \times 10^{6} \times 10^{- 5} = 8 \times 10^{- 18}N$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.