Understanding the Physics of a Proton in a Magnetic Field

This question asks us to determine the kinetic energy of a proton after it travels through a specific path within a uniform magnetic field. We are given that the proton initially has a kinetic energy $T$ and enters a region with a magnetic field $\vec{B}$ directed perpendicularly to its initial velocity $\vec{v}$. The proton travels exactly one-quarter of a circular path before exiting the field.

Key Physics Principles

To solve this, we need to consider the fundamental principles governing the motion of charged particles in magnetic fields and the concept of kinetic energy:

• Magnetic Force: A charged particle moving in a magnetic field experiences a force given by the Lorentz force equation: $\vec{F} = q(\vec{v} \times \vec{B})$. For a proton, the charge $q$ is positive ($q = e$).
• Motion in a Perpendicular Magnetic Field: When the velocity $\vec{v}$ is perpendicular to the uniform magnetic field $\vec{B}$, the magnetic force acts as a centripetal force, causing the particle to move in a circular path. The magnitude of the force is $F = evB$.
• Work and Energy: The work done by a force on an object changes its kinetic energy, according to the work-energy theorem ($\Delta K = W$). Work is calculated as $W = \int \vec{F} \cdot d\vec{r}$.
• Kinetic Energy: The kinetic energy of an object is defined as $K = \frac{1}{2}mv^2$, where $m$ is the mass and $v$ is the speed.

Step-by-Step Analysis of Proton's Motion

1. Nature of Magnetic Force: The magnetic force $\vec{F}$ is always perpendicular to the velocity $\vec{v}$ because of the cross product ($\vec{v} \times \vec{B}$).
2. Work Done by Magnetic Force: Since the force $\vec{F}$ is always perpendicular to the direction of motion (which is the direction of $\vec{v}$ and infinitesimal displacement $d\vec{r}$), the magnetic force does zero work on the proton. Mathematically, $W_{magnetic} = \int \vec{F} \cdot d\vec{r} = \int (evB) \sin(90^\circ) dr = 0$, because the angle between $\vec{F}$ and $d\vec{r}$ is always $90^\circ$.
3. Change in Kinetic Energy: According to the work-energy theorem, the change in kinetic energy ($\Delta K$) is equal to the net work done ($W_{net}$). In this case, the only force doing work is the magnetic force. Therefore, $\Delta K = W_{magnetic} = 0$.
4. Constant Kinetic Energy: Since the change in kinetic energy is zero ($\Delta K = 0$), the kinetic energy of the proton remains constant throughout its journey in the magnetic field.
5. Initial and Final Kinetic Energy: The proton enters the magnetic field with kinetic energy $K_{initial} = T$. Because the kinetic energy does not change, the kinetic energy as it exits the magnetic field ($K_{final}$) must be the same as the initial kinetic energy.

The fact that the proton travels exactly one-quarter of a circular path is information about the geometry of the motion and the specific conditions (like the field strength, velocity, and the dimensions of the magnetic field region), but it does not affect the kinetic energy itself, as the magnetic force does no work.

Conclusion on Kinetic Energy

The proton starts with kinetic energy $T$. Since the magnetic force does no work, the kinetic energy remains unchanged. Therefore, the kinetic energy of the proton as it exits the magnetic field is still $T$.