A) \[\left( \frac{M{{v}^{2}}}{R} \right)2\pi R\]
B) zero
C) BQ \[2\pi R\]
D) BQv \[2\pi R\]
Correct Answer: B
Solution :
When particle describes circular path in a magnetic field, its velocity is always perpendicular to the magnetic force.Power | \[P=F.\,v=Fv\cos \theta \] |
Here, | \[\theta ={{90}^{o}}\] |
\[\therefore \] | P = 0 |
But, | \[P=\frac{W}{t}\Rightarrow W=P.\,\,t\] |
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