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  4. A charged particle is moving on circular path with velocity v in a uniform magnetic field B, if the velocity of the charged particle is doubled and strength of magnetic field is halved, then radius becomes

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Q. A charged particle is moving on circular path with velocity $v$ in a uniform magnetic field $B$, if the velocity of the charged particle is doubled and strength of magnetic field is halved, then radius becomes

Moving Charges and Magnetism

Solution:

As $Bqv=\frac{mv^{2}}{r}$ or $r=\frac{mv}{Bq}$
According to the question, $v'=2v$ and $B' =\frac{B}{2}$
$\therefore r'=\frac{mv'}{B' q}=\frac{m \left(2v\right)}{\left(B /2\right)q}=\frac{4mv}{Bq}=4r$