Question

A particle with a charge of \(20.0 \mu \mathrm{C}\) moves along the \(x\) -axis with a speed of \(50.0 \mathrm{~m} / \mathrm{s}\). It enters a magnetic field given by \(\vec{B}=0.300 \hat{y}+0.700 \hat{z},\) in teslas. Determine the magnitude and the direction of the magnetic force on the particle.

Short answer

Question: Determine the magnitude and direction of the magnetic force on a charged particle moving with a charge of \(20.0 \mu C\), a speed of \(50.0 m/s\) in the x-direction, and experiencing a magnetic field \(\vec{B} = 0.300 \hat{y} + 0.700 \hat{z}\). Answer: The magnetic force on the charged particle has a magnitude of \(7.81 \times 10^{-4}\) N and makes an angle of \(23.2 ^{\circ}\) relative to the negative y-axis.

Step-by-step solution

Given values

We are given the following values: Charge, \(q = 20.0 \mu C = 20.0 \times 10^{-6} C\) Velocity, \(\vec{v} = 50.0 m/s \hat{x}\) Magnetic Field, \(\vec{B} = 0.300 \hat{y} + 0.700 \hat{z}\) (in teslas)

Compute the cross product

Now, we need to compute the cross product of the velocity vector \(\vec{v}\) and the magnetic field vector \(\vec{B}\). Based on the given vectors, we will have: \(\vec{v} \times \vec{B} = \begin{vmatrix} \hat{x} & \hat{y} & \hat{z} \\ 50.0 & 0 & 0 \\ 0 & 0.300 & 0.700 \end{vmatrix}\)

Simplify the cross product

We can now simplify the determinant to get the cross product of the two vectors. \(\vec{v} \times \vec{B} = 50.0(0.300\hat{z} - 0.700\hat{y}) = -35.0\hat{y}+15.0\hat{z}\)

Calculate the magnetic force

We now use the formula for the magnetic force: \(\vec{F} = q(\vec{v} \times \vec{B})\) \(\vec{F}= 20.0 \times 10^{-6}(-35.0\hat{y} + 15.0\hat{z})\) \(\vec{F}= -7.0 \times 10^{-4} \hat{y} + 3.0 \times 10^{-4} \hat{z}\) N

Find magnitude of force

To find the magnitude of the magnetic force, we will use the formula: \(F = \sqrt{F_y^2 + F_z^2}\) \(F = \sqrt{(-7.0 \times 10^{-4})^2 + (3.0 \times 10^{-4})^2} = \sqrt{6.10 \times 10^{-7}}\) \(F = 7.81 \times 10^{-4}\) N

Find direction of force

To find the direction of the magnetic force, we will calculate the angle (\(\theta\)) relative to the y-axis (negative y-axis). We can calculate the angle using the formula: \(\theta = \arctan \Big(\frac{F_z}{ |F_y|} \Big)\) \(\theta = \arctan \Big(\frac{3.0 \times 10^{-4}}{7.0 \times 10^{-4}}\Big)\) \(\theta \approx 23.2 ^{\circ}\) The magnetic force has a magnitude of \(7.81 \times 10^{-4}\) N and makes an angle of \(23.2 ^{\circ}\) relative to the negative y-axis.

Key concepts

Cross Product in Physics

Understanding the cross product is essential when delving into physics problems involving forces in magnetic fields. In physics, the cross product, also known as the vector product, is a binary operation on two vectors in a three-dimensional space. It's given by the formula \(\vec{A} \times \vec{B}\) and results in a vector that is perpendicular to both \(\vec{A}\) and \(\vec{B}\). In our particle example, we needed to take the cross product of the particle’s velocity vector and the magnetic field vector to find the direction of the force.

The process involves a determinant, formed with the unit vectors \(\hat{x}\), \(\hat{y}\), and \(\hat{z}\), and the components of the vectors involved. The cross product plays a crucial role because the magnetic force on a moving charge is always perpendicular to the direction of its velocity and the magnetic field, which is exactly what the cross product of those two vectors would represent.

Lorentz Force Law

The Lorentz force law is a cornerstone concept in electromagnetism, defining how charged particles move in electromagnetic fields. This fundamental law states that the force \(\vec{F}\) exerted on a charge \(q\) moving with a velocity \(\vec{v}\) in an electric field \(\vec{E}\) and magnetic field \(\vec{B}\) is:

\[\vec{F} = q(\vec{E} + \vec{v} \times \vec{B})\]

In cases where only the magnetic field is present, as in our example, the equation simplifies to \(\vec{F} = q(\vec{v} \times \vec{B})\). Here, \(\vec{F}\) represents the magnetic force, \(\vec{v}\) the velocity of the particle, and \(\vec{B}\) the magnetic field. It's important to remember that the Lorentz force is what keeps particles moving in circular paths in cyclotrons and is also the principle behind devices like mass spectrometers.

Charge and Magnetic Fields Interaction

When a charged particle moves through a magnetic field, it experiences a force thanks to the interaction between the charge and the field. This interaction is intimately described by the right-hand rule, which is a visual mnemonic used to understand the direction of the produced force. In our example, a positive charge moving along the x-axis and entering a magnetic field that has components in both \(y\) and \(z\) directions will experience a force in a direction that can be determined by this right-hand rule.

The magnitude of this force is proportional to the charge, the speed of the particle, and the strength of the magnetic field. But, crucially, it also depends on the angle between the velocity of the particle and the magnetic field lines. If the velocity is parallel to the field lines, no force is produced. This is because the cross product of two parallel vectors is zero.

Vector Operations in Physics

Vectors are a fundamental part of physics, representing quantities that have both a magnitude and a direction. Vector operations, including addition, subtraction, and products (dot and cross), are crucial in solving physics problems. In our exercise, we used vector notation to symbolize the direction of the particle's movement and the magnetic field and then applied the cross product to these vectors to find the magnetic force exerted on the particle.

A good grasp of vector operations is essential when dealing with forces, fields, and motions in physics. Taking the time to visualize and perform these operations step-by-step is key in understanding complex topics such as electromagnetic fields and can be incredibly helpful in interpreting the variety of scenarios that physics presents.