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solve and explain

Asked by Arushi Juyal 13th December 2017, 7:19 AM
Answered by Expert
Answer:
Force acting on charged particle begin mathsize 12px style F with rightwards arrow on top equals q open parentheses E with rightwards arrow on top space plus v with rightwards arrow on top space cross times B with rightwards arrow on top space close parentheses end style
Since there is no change in magnitude and direction for the charged particle travelled in the region of electric and magnetic fields, the net force is zero
 
begin mathsize 12px style H e n c e space E with rightwards arrow on top space plus space v with rightwards arrow on top cross times B with rightwards arrow on top space equals space 0 E with rightwards arrow on top space equals space minus space v with rightwards arrow on top cross times B with rightwards arrow on top space equals space B with rightwards arrow on top space cross times v with rightwards arrow on top space E with rightwards arrow on top space cross times space B with rightwards arrow on top space equals B with rightwards arrow on top space cross times v with rightwards arrow on top cross times stack B space with rightwards arrow on top space equals space open parentheses B with rightwards arrow on top times B with rightwards arrow on top close parentheses v with rightwards arrow on top space minus open parentheses B with rightwards arrow on top times v with rightwards arrow on top close parentheses B with rightwards arrow on top space equals space B squared v with rightwards arrow on top space left parenthesis space sin c e space B with rightwards arrow on top space a n d space v with rightwards arrow on top space a r e space p e r p e n d i c u l a r right parenthesis end style
 
hence begin mathsize 12px style v with rightwards arrow on top space equals fraction numerator E with rightwards arrow on top space cross times B with rightwards arrow on top over denominator B squared end fraction space end style

Answered by Expert 13th December 2017, 9:21 PM
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