Problem 8 - College Physics

Question

A particle with a charge of \(-2.50 \times 10^{-8} \mathrm{C}\) is moving with an instantaneous velocity of magnitude 40.0 \(\mathrm{km} / \mathrm{s}\) in the \(x y-\) plane at an angle of \(50^{\circ}\) counterclockwise from the \(+x\) axis. What are the magnitude and direction of the force exerted on this particle by a magnetic field with magnitude 2.00 T in the (a) \(-x\) direction, and (b) \(+z\) direction?

Step-by-Step Solution

Answer

(a) The force magnitude is approximately \(7.66 \times 10^{-4} \mathrm{N}\) in the \(-z\) direction. (b) The force magnitude is the same approximately \(7.66 \times 10^{-4} \mathrm{N}\) with components in the negative \(x\) and positive \(y\) directions.

Step 1: Understanding the Scenario

When a charged particle moves through a magnetic field, it experiences a force given by the Lorentz force equation: \ \( \mathbf{F} = q (\mathbf{v} \times \mathbf{B}) \) \ where \( q \) is the charge of the particle, \( \mathbf{v} \) is its velocity vector, and \( \mathbf{B} \) is the magnetic field vector.

Step 2: Determining the Velocity Vector

The velocity vector \( \mathbf{v} \) can be decomposed into components in the \( x \) and \( y \) directions based on the angle given. \ The magnitude of the velocity is \( 40000 \ \mathrm{m/s} \), and the angle is \( 50^{\circ} \). Therefore: \ \( v_x = 40000 \cos(50^{\circ})\) \ \( v_y = 40000 \sin(50^{\circ}) \) \ The velocity vector is thus \( \mathbf{v} = (40000 \cos(50^{\circ})) \hat{i} + (40000 \sin(50^{\circ})) \hat{j} \).

Step 3: Calculating the Force in the -x Direction

In this part, the magnetic field \( \mathbf{B} \) is in the \(-x\) direction, so \( \mathbf{B} = -2.00 \hat{i} \ \mathrm{T} \). \ We use the cross product \( \mathbf{F} = q (\mathbf{v} \times \mathbf{B}) \) : \ \( \mathbf{F} = (-2.50 \times 10^{-8} \mathrm{C}) \left((40000 \cos(50^{\circ}) \hat{i} + 40000 \sin(50^{\circ}) \hat{j}) \times (-2.00 \hat{i})\right) \) \ Solve the cross product: \ \( (40000 \sin(50^{\circ}) \hat{j}) \times (-2.00 \hat{i}) = -2.00 \times 40000 \sin(50^{\circ}) \hat{k} \) \ Force magnitude: \ \( F = 2.50 \times 10^{-8} \times 2.00 \times 40000 \sin(50^{\circ}) \). \ Calculate \( F \).

Step 4: Calculating the Force in the +z Direction

Here, the magnetic field \( \mathbf{B} \) is in the \(+z\) direction, so \( \mathbf{B} = 2.00 \hat{k} \ \mathrm{T} \). \ Use the cross product \( \mathbf{F} = q (\mathbf{v} \times \mathbf{B}) \) : \ \( \mathbf{F} = (-2.50 \times 10^{-8} \mathrm{C}) \left((40000 \cos(50^{\circ}) \hat{i} + 40000 \sin(50^{\circ}) \hat{j}) \times 2.00 \hat{k}\right) \) \ Solve the cross product: \ \( (40000 \cos(50^{\circ}) \hat{i} \times 2.00 \hat{k}) = 2.00 \times 40000 \cos(50^{\circ}) \hat{j} \) \ \( (40000 \sin(50^{\circ}) \hat{j} \times 2.00 \hat{k}) = -2.00 \times 40000 \sin(50^{\circ}) \hat{i} \) \ Determine direction and magnitude of \( \mathbf{F} \).

Key Concepts

magnetic field

A magnetic field is a region of space where a magnetic force is experienced by moving charged particles or by magnetic materials. It's like invisible lines of force that originate from magnets or electrical currents. The strength and direction of a magnetic field are represented by the vector \( \mathbf{B} \), and the unit of measurement is the Tesla (T).

Magnetic fields have different orientations and magnitudes based on their source and surrounding environment. In our problem, the magnetic field is given in two directions:

• \(-x\) direction: where the magnetic field vector is \(-2.00 \hat{i}\)
• \(+z\) direction: where the magnetic field vector is \(2.00 \hat{k}\)

Understanding the orientation of the magnetic field is crucial for calculating the Lorentz force, which is what a charged particle feels when moving through such a field.

charged particle

A charged particle is simply a particle with an electric charge. In physics, many particles such as electrons, protons, and ions carry a charge, which might be positive or negative. The charge is an intrinsic property of the particle, and it's measured in coulombs (C).

In the given exercise, our particle has a charge of \(-2.50 \times 10^{-8} \text{ C} \). The negative sign indicates that it is negatively charged, similar to how electrons are charged. This charge influences how it interacts with electric and magnetic fields through the Lorentz force.

When a charged particle moves through a magnetic field, the force acting on it is proportional to:

• The magnitude of the charge
• The velocity of the particle
• The strength and orientation of the magnetic field

The interplay of these aspects will determine the direction and magnitude of the force the particle experiences.

cross product

The cross product is a mathematical operation used to determine a vector that is perpendicular to two given vectors in three-dimensional space. In the context of a charged particle moving within a magnetic field, the cross product is used to calculate the Lorentz force: \( \mathbf{F} = q(\mathbf{v} \times \mathbf{B}) \).

The cross product \( \mathbf{v} \times \mathbf{B} \) is significant because:

• It determines the direction of the force using the right-hand rule.
• Only the components of \( \mathbf{v} \) and \( \mathbf{B} \) that are perpendicular contribute to the magnitude of the force.

Calculating the cross product requires identifying the vector components and their directions. In our exercise:

• For a magnetic field in the \(-x\) direction, the cross product involves components of the velocity in the \(y\) direction.
• For a magnetic field in the \(+z\) direction, the cross products involve components of the velocity in both \(x\) and \(y\) directions.

This operation, although computationally straightforward, requires careful attention to the vector directions and magnitudes to accurately find the resultant force.

velocity vector

The velocity vector \( \mathbf{v} \) of a particle is an essential concept because it describes not only how fast but also in which direction the particle is moving. Velocity has both magnitude (speed) and direction, and it's represented as a vector. In our exercise, the velocity's magnitude is given as 40.0 km/s, and its direction makes an angle of \(50^\circ\) counterclockwise from the \(+x\) axis.

To work with this vector mathematically, we need to decompose it into components along the \(x\) and \(y\) axes. Using trigonometric functions:

• \(v_x = 40000 \cos(50^\circ)\) for the \(x\)-component
• \(v_y = 40000 \sin(50^\circ)\) for the \(y\)-component

This decomposition allows us to use these components along with the direction of the magnetic field to calculate the force acting on the charged particle. Having a clear understanding of the velocity vector is crucial, as it directly affects the computation and results of the Lorentz force due to its interaction with the magnetic field.