An electron of mass $m$ and charge $q$ is travalling with a speed $v$ along a circular path of radius $r$ at right angles to a uniform of magnetic field $B.$ If speed of the electron is doubled and the magnetic field is halved, then resulting path would have a radius of

Question

Dr. Abubacker · Accepted Answer

(d) In a perpendicular magnetic field,
Magnetic force = centripetal force
$ie,\ \ Bqv = \frac{mv^{2}}{r}$
$\Rightarrow \ \ r = \frac{mv}{Br}\  \Rightarrow r \propto v^{2}$
$	herefore\ \frac{r_{2}}{r} = \frac{v_{2}^{2}}{v_{1}^{2}} = \left( \frac{2v}{v} ight)^{2} = 4$
$\Rightarrow r_{2} = 4r$

Q. An electron of mass $m$ and charge $q$ is travalling with a speed $v$ along a circular path of radius $r$ at right angles to a uniform of magnetic field $B.$ If speed of the electron is doubled and the magnetic field is halved, then resulting path would have a radius of

Explanation

Reviewed by Dr. Abubacker

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