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Asked by rbhatt16 14th July 2018, 9:57 PM
Answered by Expert
Answer:
Figure given above illustrates the informations given in the question.
Using polar coordinate system we can calculate the width d of field region as
 
d = R sin60 - R sin 30 = R[ (√3/2) - (1/2)] .......(1)
 
we know in magnetic field region a charge of q Coulomb moves in a circular path and force acting on the charge is q×B×v,
where B is magnetic field strength and v is speed of charge.
 
hence we have begin mathsize 12px style q cross times B cross times v space equals space fraction numerator m cross times v squared over denominator R end fraction space space o r space R space equals space fraction numerator m cross times v over denominator q cross times B end fraction space............. left parenthesis 2 right parenthesis end style
from (1) and (2), we have begin mathsize 12px style d space equals space fraction numerator open parentheses square root of 3 space minus space 1 close parentheses cross times m cross times v over denominator 2 cross times q cross times B end fraction end style
Answered by Expert 16th July 2018, 12:17 PM
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