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Q. Assertion : If an electron and proton enter a perpendicular magnetic field with equal momentum, then radius of curve for electron is more than that of proton.
Reason : Electron has less mass than proton.

AIIMSAIIMS 2013

Solution:

When a charged particle enters in perpendicular magnetic field than radius of curved path is given by
$r=\frac{m v}{q B}=\frac{p}{q B}$
As momentum $p$ is constant
$ \therefore r \propto \frac{1}{q}$
As $e^-$ and proton have same charge
$\frac{r_{e}^{-}}{r_{p}^{+}}=\frac{q_{e}^{-}}{q_{e}^{+}}=1$
Assertion is false, but reason is true