Question

IF A CHARGE Q HAVING MASS M MOVES PERPENDICULAR TO MAGNETIC INDUCTION FIELD B WITH VELOCITY V THEN,THE RADIUS OF CIRCULAR PATH OF MOVING CHARGE
1)MV/BQ
2)BQ/MV
3)MVBQ
4)0
EXPLANATION IS REQUIRED

Answer 1

answer : option (1) MV/QB

explanation : A charge Q having mass M moves perpendicular to magnetic field B with velocity V. then, The magnetic force is perpendicular to its the velocity, and so velocity changes in direction but not magnitude. Uniform circular motion results.

see figure, here it is clearly shown that magnetic field perpendicular to its velocity. due to this velocity changes its direction but not magnitude.

so, motion of charged particle is like  uniform circular motion.

here, magnetic force is balanced by centripetal force.

i.e., BQV = MV²/R

or, R = MV/QB

hence , radius of circular path of moving charge is MV/QB

[ note : as you know, magnetic force = q(V × B) because velocity and magnetic field are perpendicular so, sin90° = 1 hence, magnitude of magnetic force = BQVsin90° = BQV ]