Charged Particle In Magnetic Field: Problems & Solutions

Let's dive into the fascinating world of how charged particles behave when they encounter a magnetic field! We'll explore different scenarios and break down the physics behind them. Get ready, because we're about to make some seriously cool discoveries!

Understanding the Basics

Before we jump into specific problems, let's quickly recap the fundamental principles at play. The key player here is the Lorentz force. This force acts on a charged particle moving in a magnetic field, and its magnitude is given by:

F = qvBsin(θ)

Where:

• F is the magnitude of the force.
• q is the magnitude of the charge of the particle.
• v is the velocity of the particle.
• B is the magnetic field strength.
• θ is the angle between the velocity vector and the magnetic field vector.

The direction of this force is perpendicular to both the velocity and the magnetic field, as determined by the right-hand rule. This means the force will cause the particle to change direction, but not speed – leading to some interesting trajectories.

Right-Hand Rule

The right-hand rule is your best friend when dealing with magnetic forces. Here's how to use it:

1. Point your fingers in the direction of the velocity v.
2. Curl your fingers towards the direction of the magnetic field B.
3. Your thumb will then point in the direction of the force F on a positive charge. If the charge is negative, the force is in the opposite direction.

Scenario 1: Particle Moving Perpendicular to the Magnetic Field

Problem Statement

A particle with charge q and mass m is moving with a velocity v perpendicular to a uniform magnetic field B. Describe the motion of the particle and determine the radius of its trajectory.

Solution

When a charged particle enters a magnetic field perpendicularly, the magnetic force acts as a centripetal force, causing the particle to move in a circle. The magnetic force continuously changes the direction of the velocity, keeping the speed constant. The key here is to equate the magnetic force with the centripetal force:

qvB = mv²/r

Where:

• r is the radius of the circular path.

Solving for r, we get:

r = mv / (qB)

This equation tells us that the radius of the circular path is directly proportional to the particle's momentum (mv) and inversely proportional to the charge and the magnetic field strength. A stronger magnetic field or a larger charge will result in a smaller radius, while a higher momentum will lead to a larger radius.

Diagram

Imagine a positive charge moving to the right and a magnetic field pointing into the page. The right-hand rule tells us that the force will be upwards, causing the particle to curve upwards. As the particle continues to move, the force will always be perpendicular to its velocity, resulting in a circular path. Draw a circle in your notebook, indicate the direction of v, B, and F at different points along the circle.

Scenario 2: Particle Moving at an Angle to the Magnetic Field

Problem Statement

A particle with charge q and mass m is moving with a velocity v at an angle θ to a uniform magnetic field B. Describe the motion of the particle.

Solution

This scenario is a bit more complex, but super interesting! We need to break down the velocity into two components:

• v parallel: The component of velocity parallel to the magnetic field (v cos θ).
• v perpendicular: The component of velocity perpendicular to the magnetic field (v sin θ).

The v parallel component experiences no force from the magnetic field, so the particle continues to move with this velocity along the direction of the field. The v perpendicular component, however, experiences a force just like in Scenario 1, causing the particle to move in a circle in a plane perpendicular to the magnetic field.

The combination of these two motions results in a helical path. The particle spirals around the magnetic field lines. The radius of the helix is determined by the v perpendicular component:

r = (m v sin θ) / (qB)

The pitch of the helix (the distance between successive turns) is determined by the v parallel component:

p = v cos θ * T

Where T is the period of the circular motion:

T = 2πm / (qB)

So, the pitch is:

p = (2πm v cos θ) / (qB)

Diagram

Imagine a magnetic field pointing to the right. Now, visualize a particle moving diagonally. The particle will spiral around the magnetic field lines, moving to the right while simultaneously moving in a circle. This creates a helix shape. Draw the magnetic field lines, the helical path of the particle, and indicate the v parallel and v perpendicular components of the velocity. It's essential to visualize this in 3D!

Scenario 3: A Velocity Selector

Problem Statement

A beam of particles with different velocities enters a region with both electric field E and magnetic field B, which are perpendicular to each other and to the velocity of the particles. Determine the velocity of the particles that will pass through the region undeflected.

Solution

In this scenario, we have two forces acting on the charged particles: the electric force and the magnetic force. For a particle to pass through undeflected, these forces must be equal and opposite.

The electric force is:

FE = qE

The magnetic force is:

FB = qvB

For no deflection, FE = FB, so:

qE = qvB

Solving for v, we get:

v = E / B

This means that only particles with a velocity equal to the ratio of the electric field strength to the magnetic field strength will pass through undeflected. This is the principle behind a velocity selector, which is used to select particles with a specific velocity from a beam of particles with a range of velocities.

Diagram

Draw three parallel plates. The top plate is positive, and the bottom plate is negative, creating an electric field pointing downwards. Now, add a magnetic field pointing into the page. A positive charge moving to the right will experience a downward electric force and an upward magnetic force. For the particle to go straight, those forces must balance. Make sure to label E, B, v, FE, and FB in your diagram.

Scenario 4: Mass Spectrometer

Problem Statement

Ions of the same charge but different masses are accelerated through an electric potential and then enter a region with a uniform magnetic field. Describe how the magnetic field separates the ions based on their mass and derive an expression for the mass of the ions in terms of the radius of their path in the magnetic field.

Solution

First, the ions are accelerated through an electric potential V. The kinetic energy gained by the ions is equal to the electric potential energy lost:

KE = qV = 1/2 mv²

Solving for v, we get:

v = √(2qV / m)

Now, the ions enter the magnetic field. As we saw in Scenario 1, they move in a circle with a radius:

r = mv / (qB)

Substituting the expression for v from above, we get:

r = m √(2qV / m) / (qB) = √(2mV / q) / B

Solving for m, we get:

m = (qB²r²) / (2V)

This equation shows that the radius of the circular path is directly proportional to the square root of the mass of the ion. Therefore, ions with different masses will follow different paths in the magnetic field, allowing us to separate them and determine their masses. This is the principle behind a mass spectrometer.

Diagram

Draw a region where ions are being accelerated by an electric potential. Then, draw a region with a magnetic field. Show ions of different masses following different circular paths in the magnetic field. The heavier ions will have larger radii. Be sure to label V, B, and the radii of the different paths.

Scenario 5: The Earth's Magnetic Field

Problem Statement

Charged particles from the sun (solar wind) are constantly bombarding the Earth. Explain how the Earth's magnetic field protects us from these harmful particles and what phenomena result from this interaction.

Solution

The Earth's magnetic field acts as a giant shield, deflecting most of the charged particles from the solar wind. These particles, mostly electrons and protons, are deflected by the Lorentz force. Instead of reaching the surface of the Earth, they are forced to spiral around the magnetic field lines. The Earth's magnetic field lines converge at the poles, and as the particles spiral along these lines, they are funneled towards the polar regions.

When these charged particles collide with atoms and molecules in the Earth's atmosphere (mainly oxygen and nitrogen), they excite these atoms. When the excited atoms return to their ground state, they emit light, creating the beautiful auroras (Northern and Southern Lights). The color of the aurora depends on the type of atom being excited. Oxygen typically emits green and red light, while nitrogen emits blue and purple light.

Diagram

Draw a diagram of the Earth surrounded by its magnetic field lines. Show charged particles from the sun being deflected by the field lines and spiraling towards the poles. Indicate the regions where the auroras occur. Label the Earth, the magnetic field lines, the solar wind, and the auroras.

Scenario 6: Magnetic Force on a Current-Carrying Wire

Problem Statement

A straight wire of length L carries a current I in a uniform magnetic field B. The wire makes an angle θ with the magnetic field. Determine the magnitude and direction of the magnetic force on the wire.

Solution

The magnetic force on a current-carrying wire is given by:

F = I L B sin(θ)

Where:

• F is the magnitude of the force.
• I is the current in the wire.
• L is the length of the wire.
• B is the magnetic field strength.
• θ is the angle between the wire and the magnetic field.

The direction of the force is perpendicular to both the wire and the magnetic field, as determined by the right-hand rule. Point your fingers in the direction of the current, curl them towards the direction of the magnetic field, and your thumb will point in the direction of the force.

Diagram

Draw a straight wire of length L carrying a current I in a magnetic field B. Indicate the angle θ between the wire and the magnetic field. Use the right-hand rule to determine the direction of the force F on the wire. Label I, L, B, θ, and F in your diagram.

Scenario 7: Magnetic Torque on a Current Loop

Problem Statement

A rectangular loop of wire with sides a and b carries a current I in a uniform magnetic field B. The normal to the loop makes an angle θ with the magnetic field. Determine the magnetic torque on the loop.

Solution

The magnetic torque on a current loop is given by:

τ = I A B sin(θ)

Where:

• τ is the magnitude of the torque.
• I is the current in the loop.
• A is the area of the loop (A = ab).
• B is the magnetic field strength.
• θ is the angle between the normal to the loop and the magnetic field.

We can also express the torque in terms of the magnetic dipole moment μ of the loop:

μ = I A

So, the torque becomes:

τ = μ B sin(θ)

The torque tends to align the magnetic dipole moment of the loop with the magnetic field.

Diagram

Draw a rectangular loop of wire with sides a and b carrying a current I in a magnetic field B. Indicate the angle θ between the normal to the loop and the magnetic field. Show the direction of the torque τ on the loop. Label I, A, B, θ, τ, and the normal vector in your diagram.

Conclusion

So, there you have it! We've explored a variety of scenarios involving charged particles and magnetic fields. From simple circular motion to complex helical paths and the workings of devices like velocity selectors and mass spectrometers, the interaction between charge and magnetism is a fundamental and fascinating aspect of physics. By understanding these concepts and practicing with different problems, you'll be well on your way to mastering electromagnetism! Keep exploring, keep questioning, and keep learning!