Kerala KEAM 5 June 2024 Solved Paper

The magnetic dipole moment $\mu$ of an electron moving in a circular orbit is given by: $$\mu = I \cdot A$$where $I$ is the current due to the motion of the electron, and $A$ is the area of the orbit.For a revolving electron, the current $I$ is given by the charge per unit time, which is: $$I = \frac{e}{T}$$where $T$ is the time period of the electron’s revolution. The area $A$ of the orbit is given by: $$A = \pi r^2$$where $r$ is the radius of the orbit.Now, the angular momentum $L$ of the electron is given by: $$L = mvr$$where $v$ is the velocity of the electron and $r$ is the radius of the orbit.Thus, the ratio between the magnetic dipole moment and angular momentum is: $$\frac{\mu}{L} = \frac{\frac{e}{T} \cdot \pi r^2}{mvr}$$We know that $v = \frac{2\pi r}{T}$, so: $$\frac{\mu}{L} = \frac{\frac{e}{T} \cdot \pi r^2}{m \cdot \frac{2\pi r}{T} \cdot r} = \frac{e}{2m}$$Thus, the correct answer is option (C), $\frac{e}{2m}$. Quick Tip: In problems involving revolving electrons and magnetic dipole moment, remember that the ratio of the magnetic dipole moment to the angular momentum is $\frac{e}{2m}$, derived from the relationship between the current, area, and angular momentum.