AIPMT PRELIMS 2001 Physics Cyclotron Motion MCQ Question

When an electron with mass 'm' and charge 'e' enters a uniform magnetic field 'B' perpendicularly, it experiences a magnetic force that acts as a centripetal force, causing it to move in a circular path. The magnetic force is given by: $F = e v B$ where 'v' is the velocity of the electron. This force provides the centripetal acceleration, which can be expressed as: $F = \frac{m v^2}{r}$ Equating these two expressions for force, we have: $e v B = \frac{m v^2}{r}$ From this, we can derive the radius of the circular path as: $r = \frac{m v}{e B}$ The frequency of the electron's motion in the magnetic field (also known as cyclotron frequency) is given by: $f = \frac{v}{2 \pi r}$ Substituting the expression for 'r', we get: $f = \frac{v e B}{2 \pi m v} = \frac{e B}{2 \pi m}$ Thus, the correct answer is C) $\frac{e B}{2 \pi m}$.

For the other options:

• A) $\frac{eE}{qVB}$ does not relate to the frequency of motion in a magnetic field.
• B) $\frac{2 \pi m}{eB}$ is incorrect as it implies an inverse relationship with frequency.
• D) $\frac{2m}{eBE}$ also does not represent the frequency correctly and involves energy incorrectly. Therefore, C is the only correct choice.