Q. If a particle of charge $10^{- 12}$ C moving along the $x$-direction with a velocity of $10^{5}\ ms^{- 1}$ experience a force of $10^{- 10}$ N in $y$-direction due to magnetic field, then the minimum value of magnetic field is
A) $6.25 \times 10^{3}\ T\ in\ z$-direction
B) $10^{- 15}\ T\ in\ z$-direction
C) $6.25 \times 10^{- 3}\ T\ in\ z$-direction
D) $10^{- 3}\ T\ in\ z$-direction
Explanation
(d) $B = \frac{F}{q\ v\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.