• # question_answer A charged particle of mass m and charge q is projected into a uniform magnetic field of induction B with speed v which is perpendicular to B. The width of the magnetic field is d. The impulse imparted to fee particle by the field is $\left( d<<\frac{mv}{qB} \right)$ A) $qBv$                          B) $\frac{mv}{qB}$ C) $qBd$              D) $\frac{2\,m{{v}^{2}}}{qB}$

[C]  q more on circular path in B. Whose radius $=\frac{mv}{qB}=r$ In question width of magnetic field region is very small than r. so charge q will exit from B on opposite side of region path of q is shown in figure.
 As $d<\,<\,<\frac{mv}{qB}$ Using $\sin \theta =\frac{d}{r}$ Impulse is change in momentum
 Where $r=\frac{mv}{qB}$ \[\therefore d<<