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  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)

Syed SAAD , 6 Years ago
Grade 12
anser 2 Answers
ridam agarwal
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.
   
 
Last Activity: 5 Years ago
Vikas TU
Dear student 
Please refer 
 
Last Activity: 5 Years ago
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