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# A particle of mass 0.5 g and charge 2.5×10-8C is moving with velocity 6×10 4m/s what should be the minimum value of magnetic field acting on it so that the particle is able to move in a straight line(g=9.8m/s2)

ridam agarwal
26 Points
one year ago
q=2.5*10-8C
v=6*104m/s
since when magnetic field is not present then only gravitational force is act on the particle in vertical downward direction  but according to the question particle should be move in a straight line.
so to balance this vertical downward force there should be magnetic force in vertically upward direction.
and we know if magnetic field is B then magnetic force is given by
= q(v*B)
and it should be equal to
= mg

so
q(v*B)=mg
2.5*10-8(6*104*B)=0.5*10-3*9.8
on solving this equation we get
B=3.2666 tesla
and this is minimum value of magnetic field because magnetic force is act only in upward direction.

Vikas TU
14149 Points
11 months ago