Question

a proton and an electron travelling along parallel path enter a region of uniform magnetic field, acting perpendicular to this path. which of them will move in circular path with high frequency

Answer 1

As we know that f=qB/2(\pi)m


So mass is inversely proportional to frequency.

Since electron has lower mass than a proton, So it will have a higher frequency.

Answer 2

Answer:

The electron will move in a circular path with a more increased frequency.

Explanation:

The mass of an electron exists low as an approximated proton. Hence when both join into the uniform magnetic region, the electron will move in a circular path with a more increased frequency in the opposing direction to the current.

The frequency of a charged particle exists shown by

$v &=\frac{q B}{2 \pi m}$

$\Rightarrow v & \propto \frac{1}{m}\end{aligned}$

Now, as we understand that the mass of a proton exists more significant than that of an electron, i.e. $m_{p} > m_{e}$ and frequency exists inversely proportional to the mass, therefore the electrons will move in a circular path with higher frequency.

Therefore, the electron will move in a circular path with a more increased frequency.

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