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Grade: 12
        
  1. 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)
one year ago

Answers : (2)

ridam agarwal
26 Points
							
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.
   
 
2 months ago
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