Q. If a particle of charge $10^{- 12}\ C$ moving along the $x$-direction with a velocity of $10^{5}\ m/s$ experience a force of $10^{- 10}\ N$ in $y$-direction due to magnetic field, then the minimum value of magnetic field is
A) $6.25 \times 10^{3}\ T\ in\ z - direciton\ $
B) $10^{- 15}\ T\ in\ z - direction$
C) $6.25 \times 10^{- 3}\ T\ in\ z - direction$
D) $10^{- 3}\ T\ in\ z - direciton$
Explanation
(d) The minimum value of magnetic field
$B = \frac{F}{qv\sin{90{^\circ}}}$
$\ \ \ \ = \frac{10^{- 10}}{10^{- 12} \times 10^{5}} = 10^{- 3}\ T\ in\ z - direction\ $
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.