Problem 66 - Fundamentals of Physics

Question

A proton of charge \(+e\) and mass \(m\) enters a uniform magnetic field \(\vec{B}=B \hat{i}\) with an initial velocity \(\vec{v}=v_{0 x} \hat{i}+v_{0 y} \hat{j} .\) Find an expression in unit-vector notation for its velocity \(\vec{v}\) at any later time \(t\).

Step-by-Step Solution

Answer

The velocity is \( \vec{v}(t) = v_{0x} \hat{i} + v_{0y} \cos \left( \frac{eB}{m} t \right) \hat{j} + v_{0y} \sin \left( \frac{eB}{m} t \right) \hat{k} \).

Step 1: Understanding the motion

When a charged particle moves in a magnetic field, it experiences a magnetic force given by the Lorentz force equation \( \vec{F} = q(\vec{v} \times \vec{B}) \). In this scenario, the magnetic field \( \vec{B} \) is along the x-axis, \( \vec{B} = B \hat{i} \), so it affects only the components of velocity that are perpendicular to it.

Step 2: Calculating the force components

The initial velocity of the proton \( \vec{v} = v_{0x} \hat{i} + v_{0y} \hat{j} \) has components along both the x-axis and y-axis. The magnetic force is calculated as follows:\[ \vec{F} = +e \left( (v_{0x} \hat{i} + v_{0y} \hat{j}) \times B \hat{i} \right) \]Applying the cross product:\[ \vec{F} = +eB(v_{0y} \hat{k}) \]The force is perpendicular to both \( \vec{v} \) and \( \vec{B} \), affecting only the y-direction initially.

Step 3: Determining the motion type

The force causes the particle to undergo circular motion in a plane perpendicular to the magnetic field. The velocity component along the x-direction remains constant because \( \vec{F} \) has no x-component. The y and z-components will change over time due to the circular motion.

Step 4: Writing the velocity equations

In uniform circular motion, the magnitude of the velocity doesn't change, but its direction does. The proton will move in a circle of radius \( r = \frac{mv_{0y}}{eB} \), with periods of circular motion:- Velocity in the x-direction remains \( v_{x}(t) = v_{0x} \) - Velocity in the y-direction and z-components changes as:\[ v_{y}(t) = v_{0y} \cos \left( \frac{eB}{m} t \right) \] \[ v_{z}(t) = v_{0y} \sin \left( \frac{eB}{m} t \right) \]

Step 5: Final expression in unit-vector notation

The velocity \( \vec{v}(t) \) at any later time \( t \) considering the circular motion of the particle in the y and z plane while keeping the x component constant is:\[ \vec{v}(t) = v_{0x} \hat{i} + v_{0y} \cos \left( \frac{eB}{m} t \right) \hat{j} + v_{0y} \sin \left( \frac{eB}{m} t \right) \hat{k} \]

Key Concepts

Circular Motion

Circular motion occurs when an object moves in a circular path due to a centripetal force acting on it. In the case of a charged particle like our proton, the centripetal force is provided by the magnetic Lorentz force. This force causes the particle to move in a circular orbit within a plane that is perpendicular to the magnetic field.

The key element here is that while the particle is moving in circular motion, the magnitude of its velocity remains constant. Only the direction of the velocity changes as it moves along the circle. This is due to the nature of centripetal acceleration, which always acts perpendicular to the direction of the particle's instantaneous velocity.

• The radius of this circular path can be determined using the formula: \( r = \frac{mv_{0y}}{eB} \). Here, \( m \) is the mass of the proton, \( v_{0y} \) is the initial velocity component perpendicular to the magnetic field, \( e \) is the charge of the proton, and \( B \) is the magnetic field strength.
• The proton's trajectory forms a circle in the y-z plane because these are the directions perpendicular to the magnetic field along the x-axis.
• The velocity vector constantly changes but the speed remains constant due to the conservative nature of magnetic forces.

Magnetic Field

Magnetic fields are invisible fields that exert forces on moving electric charges, such as our proton. These fields can be defined in terms of their direction and strength, which dictate how they interact with charged particles. In this scenario, our magnetic field \( \vec{B} \) is directed along the x-axis, specifically in the direction of \( \hat{i} \).

The impact of a magnetic field on a charged particle is determined by the Lorentz force, which is given by the equation \( \vec{F} = q(\vec{v} \times \vec{B}) \), where:

• \( \vec{v} \) is the velocity of the charged particle,
• \( \vec{B} \) is the magnetic field, and
• \( q \) is the charge of the particle.

The force \( \vec{F} \) acts perpendicularly to both the magnetic field and the velocity of the particle, causing circular motion.

In this example, the magnetic force impacts the y-component of velocity of the proton, making it enter a circular motion in the y-z plane, while the x-component of velocity remains unchanged. The magnetic force is essentially what keeps the proton spinning around in its circular path.

Velocity in Magnetic Field

The behavior of a charged particle's velocity when it enters a magnetic field is fascinating and is governed by the magnetic Lorentz force. Upon entry, the components of velocity that are not aligned with the magnetic field experience a force, causing them to rotate perpendicularly to the magnetic field direction.

In our problem, initially, the proton has a velocity \( v_{0x} \) along the x-axis and \( v_{0y} \) along the y-axis. Given that the magnetic field only affects components of velocity that are perpendicular to it, this means:

• The velocity in the x-direction (\( v_{x}(t) \)) remains constant at \( v_{0x} \).
• The velocity along the y-direction and z-components transform over time due to the circular pattern in the y-z plane:
• \( v_{y}(t) = v_{0y} \cos \left( \frac{eB}{m} t \right) \)
• \( v_{z}(t) = v_{0y} \sin \left( \frac{eB}{m} t \right) \)

These equations describe simple harmonic motion, where \( v_{0y} \) is the amplitude of oscillating components in the circular path.

Overall, while the proton's speed does not change, its velocity vector changes direction, wrapping it into a circle in the presence of the magnetic field.