Understanding Charged Particle Motion in a Uniform Magnetic Field

When a charged particle moves in a uniform magnetic field, it experiences a magnetic force known as the Lorentz force. If the particle's velocity is perpendicular to the magnetic field direction, this force acts perpendicular to both the velocity and the magnetic field. In the case of a uniform magnetic field, this force provides the necessary centripetal force for the particle to move in a circular path.

Calculating the Radius of the Circular Path

Let the charge of the particle be $q$, its mass be $m$, its speed be $v$, and the magnitude of the uniform magnetic field be $B$. The magnitude of the Lorentz force acting on the particle when its velocity is perpendicular to the magnetic field is given by:

$\text{F}_L = |q|vB$

For the particle to move in a circle of radius $r$ with constant speed $v$, a centripetal force is required, given by:

$\text{F}_c = \frac{mv^2}{r}$

Since the Lorentz force provides the centripetal force, we can equate them:

$|q|vB = \frac{mv^2}{r}$

We can rearrange this equation to find the radius $r$:

$r = \frac{mv^2}{|q|vB}$

Since $v \neq 0$ (the particle is moving), we can cancel one $v$ from the numerator and denominator:

$r = \frac{mv}{|q|B}$

This equation shows that the radius of the circular path depends on the mass $m$, speed $v$, charge magnitude $|q|$, and magnetic field strength $B$.

Determining the Time Period of Revolution

The time period $T$ of one complete revolution is the time taken to travel the circumference of the circle ($2\pi r$) at a constant speed $v$. The formula for the time period is:

$T = \frac{\text{Distance}}{\text{Speed}} = \frac{2\pi r}{v}$

Now, substitute the expression for $r$ that we derived:

$T = \frac{2\pi \left(\frac{mv}{|q|B}\right)}{v}$

We can see that the speed $v$ appears in both the numerator and the denominator. As long as $v \neq 0$, we can cancel $v$:

$T = \frac{2\pi m}{|q|B}$

Analysis of the Time Period Formula

The final formula for the time period of the charged particle's revolution is $T = \frac{2\pi m}{|q|B}$. Let's examine the factors in this formula:

• $2\pi$: A mathematical constant.
• $m$: The mass of the particle. This is a constant property of the specific particle.
• $|q|$: The magnitude of the charge of the particle. This is also a constant property of the particle.
• $B$: The magnitude of the uniform magnetic field. The problem states the field is uniform, meaning $B$ is constant throughout the region.

The formula for $T$ only contains constants related to the particle ($m, |q|$) and the magnetic field ($B$). It does not contain the speed $v$ of the particle or the radius $r$ of its path.

Conclusion on Time Period Dependence

Based on the derived formula $T = \frac{2\pi m}{|q|B}$, the time period of revolution for a charged particle moving in a uniform magnetic field with its velocity perpendicular to the field is independent of the particle's speed ($v$) and the radius ($r$) of its circular path. This is often referred to as the cyclotron period.

Let's consider the given options:

• Depends on v and not on r: Incorrect, as $T$ does not depend on $v$.
• Is independent of both v and r: Correct, as $T$ does not depend on $v$ or $r$.
• Depends on r and not on v: Incorrect, as $T$ does not depend on $r$.
• Depends on both v and r: Incorrect, as $T$ does not depend on either $v$ or $r$.

Therefore, the time period of the revolution of the particle is independent of both its speed $v$ and the radius $r$ of its circular path.

Revision Table: Charged Particle in Magnetic Field

Additional Information on Charged Particle Motion

This independence of the time period (and also the angular frequency $\omega = |q|B/m$) from the speed and radius is a key feature of charged particle motion in a uniform magnetic field when the velocity is perpendicular to the field. This principle is fundamental to the operation of devices like the cyclotron, which accelerates charged particles.

If the charged particle's velocity is not perpendicular to the magnetic field, the motion is a helix. The velocity component parallel to the field remains constant, while the perpendicular component causes circular motion. The time period of the circular motion (and thus the pitch of the helix) is still given by $T = \frac{2\pi m}{|q|B}$, independent of the speed components.