Question

The force acting on a moving charged particle in a magnetic field \(\mathbf{B}\) is \(\mathbf{F}=q(\mathbf{v} \times \mathbf{B})\) where \(q\) is the electric charge of the particle, and \(\mathbf{v}\) is its velocity. Suppose that a particle moves in the \((x, y)\) plane with a uniform \(\mathbf{B}\) in the \(z\) direction. Assuming Newton's second law, \(m d \mathbf{v} / d t=\mathbf{F},\) show that the force and velocity are perpendicular and that both have constant magnitude. Hint: Find \((d / d t)(\mathbf{v} \cdot \mathbf{v}).\)

Short answer

The force and velocity are perpendicular, and both have constant magnitude.

Step-by-step solution

Write the given formula

The force acting on a moving charged particle in a magnetic field is given by \ \(\mathbf{F} = q(\mathbf{v} \times \mathbf{B}) \) where \( q \) is the electric charge and \( \mathbf{v} \) is the velocity.

Newton's Second Law

According to Newton's second law, \ \( m \dfrac{d \mathbf{v}}{dt} = \mathbf{F} \). Substituting the given force, we get \ \( m \dfrac{d \mathbf{v}}{dt} = q(\mathbf{v} \times \mathbf{B}) \).

Differentiate the velocity dot product

Consider the quantity \( \mathbf{v} \cdot \mathbf{v} \). Differentiating with respect to time, \ \( \dfrac{d}{dt}(\mathbf{v} \cdot \mathbf{v}) = 2 \mathbf{v} \cdot \dfrac{d \mathbf{v}}{dt} \).

Substitute Newton's law

Substitute \( \dfrac{d \mathbf{v}}{dt} \) from Newton's law into the differentiated dot product: \ \( \dfrac{d}{dt}(\mathbf{v} \cdot \mathbf{v}) = 2 \mathbf{v} \cdot \dfrac{q}{m} (\mathbf{v} \times \mathbf{B}) \).

Evaluate the cross product

The dot product of a vector with its cross product with another vector is zero: \ \( \mathbf{v} \cdot (\mathbf{v} \times \mathbf{B}) = 0 \). Therefore, \ \( \dfrac{d}{dt}(\mathbf{v} \cdot \mathbf{v}) = 0 \).

Conclude constant magnitude

Since the time derivative of \( \mathbf{v} \cdot \mathbf{v} \) is zero, \( \mathbf{v} \cdot \mathbf{v} \) is constant. Therefore, the magnitude of velocity \( \lVert \mathbf{v} \rVert \) remains constant.

Perpendicular force and velocity

The force \( \mathbf{F} \) is given by \( q(\mathbf{v} \times \mathbf{B}) \). Since a cross product is always perpendicular to the plane containing the two vectors, \( \mathbf{v} \times \mathbf{B} \) is perpendicular to \( \mathbf{v} \). Thus, \( \mathbf{F} \) is perpendicular to \( \mathbf{v} \).

Key concepts

Lorentz Force

The Lorentz force is the force exerted on a charged particle moving through an electric and magnetic field. The formula for the magnetic component of the Lorentz force is \(\textbf{F} = q(\textbf{v} \times \textbf{B})\). Here:

• \textbf{F} is the force experienced by the particle.
• \textbf{q} is the charge of the particle.
• \textbf{v} is the velocity of the particle.
• \textbf{B} is the magnetic field.

The cross product \( \textbf{v} \times \textbf{B} \) indicates that the direction of \(\textbf{F}\) is perpendicular to both \(\textbf{v}\) and \(\textbf{B}\). This orthogonal relationship can be demonstrated using the right-hand rule: point the fingers of your right hand in the direction of \(\textbf{v}\), curl them toward \(\textbf{B}\), and your thumb will point in the direction of \( \textbf{F} \). Understanding this is key to why the force does not alter the speed of the particle but rather its direction.

Cross Product

The cross product is a mathematical operation used to find a vector that is perpendicular to two given vectors. In the context of the Lorentz force, \(\textbf{F} = q(\textbf{v} \times \textbf{B})\), the cross product is crucial:

• It determines the direction of \(\textbf{F}\).
• It ensures \(\textbf{F}\) is orthogonal to both \(\textbf{v}\) and \(\textbf{B}\).
• It affects the path of the charged particle without changing its speed.

The magnitude of \( \textbf{v} \times \textbf{B} \) is given by \(|\textbf{v}||\textbf{B}|\text{sin}(\theta)\), where \(\theta\) is the angle between \(\textbf{v}\) and \(\textbf{B}\). Therefore, if \(\textbf{v}\) and \(\textbf{B}\) are perpendicular, the force is maximized, influencing how the particle curves in a circular path or helical motion, depending on other conditions in the system.

Newton's Second Law

Newton's second law of motion states that \(\textbf{F} = ma\), where force is equal to mass times acceleration. For a charged particle under a magnetic force, we use: \(\textbf{F} = m \frac{d\textbf{v}}{dt} = q(\textbf{v} \times \textbf{B})\). This connection means:

• The acceleration of the particle is due to the magnetic force.
• Applying vector differentiation, \(\frac{d\textbf{v}}{dt}\) is aligned with the force.
• Since \(\textbf{F}\) and \(\textbf{v}\) are perpendicular, the force redirects but doesn't change the magnitude of \(\textbf{v}\).

Consequently, the particle's speed (\textbf{|v|}) remains constant, while its direction changes, creating a spiral or circular trajectory.

Vector Differentiation

Vector differentiation involves differentiating vector functions with respect to time. For our context, we assess \(\textbf{v} \cdot \ \textbf{v}\):

• The dot product \(\textbf{v} \cdot \ \textbf{v}\) gives \( \textbf{|v|}^2\).
• Taking the time derivative, we get \(\frac{d}{dt}(\textbf{v} \cdot \ \textbf{v}) = 2 \textbf{v} \cdot \ \frac{d \textbf{v}}{dt}\).
• Substitute \(\frac{d \textbf{v}}{dt} = \frac{q}{m} (\textbf{v} \cdot \ \textbf{B})\): \(\frac{d}{dt}(\textbf{v} \cdot \ \textbf{v}) = 2 \textbf{v} \cdot \ \frac{q}{m} (\textbf{v} \cdot \ \textbf{B})\).

Since \(\textbf{v}\) dotted with its cross product with \(\textbf{B}\) is zero: \(\textbf{v} \cdot \ (\textbf{v} \times \ \textbf{B}) = 0\). This implies \(\frac{d}{dt} (\textbf{|v|}^2) = 0\), so \(\textbf{|v|}\) remains constant, showing that the magnitude of the particle's velocity does not change over time.