Question

The magnitude of the magnetic force on a particle with charge \(-2 e\) moving with speed \(v=1.0 \cdot 10^{5} \mathrm{~m} / \mathrm{s}\) is \(3.0 \cdot 10^{-18} \mathrm{~N}\). What is the magnitude of the magnetic field component perpendicular to the direction of motion of the particle?

Short answer

Answer: The magnitude of the magnetic field component perpendicular to the direction of motion of the particle is approximately \(9.375 \times 10^{-5}\) T.

Step-by-step solution

Write down the given information

We are given that: - the particle has a charge \(q = -2e\), - the speed of the particle is \(v = 1.0 \cdot 10^5\) m/s, - the magnitude of the magnetic force is \(F_B = 3.0 \cdot 10^{-18}\) N.

Determine the angle between the velocity vector and the magnetic field vector

In this case, we are looking for the component of the magnetic field that is perpendicular to the direction of motion of the particle, so the angle \(\theta = 90^\circ\).

Use the formula for magnetic force to find the magnitude of the magnetic field

The formula for the magnetic force on a moving charged particle is: \(F_B = |q|(vB\sin{\theta})\) Plug in the known values: \(3.0 \cdot 10^{-18} \text{ N} = |-2e|(1.0 \cdot 10^5 \text{ m/s})(B\sin{90^\circ})\) We know that \(e = 1.6 \times 10^{-19} \text{ C}\) and \(\sin{90^\circ} = 1\). Therefore, the equation becomes: \(3.0 \cdot 10^{-18} \text{ N} = |-2(1.6 \times 10^{-19} \text{ C})|(1.0 \cdot 10^5 \text{ m/s})(B)(1)\)

Solve for the magnetic field \(B\)

Rearrange the formula to solve for \(B\): \(B = \frac{3.0 \times 10^{-18} \text{ N}}{-2(1.6 \times 10^{-19} \text{ C})(1.0 \times 10^5 \text{ m/s})}\) Now, compute the value of \(B\): \(B \approx 9.375 \times 10^{-5} \text{ T}\)

State the final answer

The magnitude of the magnetic field component perpendicular to the direction of motion of the particle is approximately \(9.375 \times 10^{-5}\) T.

Key concepts

Magnetic Force

The concept of magnetic force is pivotal when understanding how charged particles interact with magnetic fields. When a charged particle, such as an electron or proton, moves through a magnetic field, it experiences a force that is perpendicular to both its velocity and the magnetic field. This force is known as the magnetic force.

• This interaction does not do work on the particle since the force is always perpendicular to the velocity of the particle.
• Instead, it alters the direction of the particle's velocity while maintaining its speed.

The formula for magnetic force (\(F_B\) ) is given by\[ F_B = |q|vB\sin{\theta} \] where \(|q|\) is the magnitude of the charge, \(v\) is the velocity of the particle, \(B\) is the magnetic field strength, and \(\theta\) is the angle between the velocity and the magnetic field lines. Understanding this force is key to analyzing the motion of charged particles in various fields, such as in particle accelerators or Earth's magnetic field.

Charged Particle Motion

Charged particle motion in a magnetic field is governed by the interplay between the particle's velocity and the magnetic force acting on it. The resulting motion is typically a circular or helical path, due to the perpendicular nature of the magnetic force to the motion of the charge.

• If a particle's velocity is perpendicular to the magnetic field lines, the particle moves in a circular path.
• If the velocity vector has a component along the magnetic field, the path becomes helical.

The centripetal force necessary for circular motion is supplied by the magnetic force:\[ F_B = \frac{mv^2}{r} \]where \(m\) is the mass of the particle, \(v\) is the speed, and \(r\) is the radius of the circular path. This equation shows how the magnetic field influences the particle's trajectory, allowing us to predict or control the path in various applications.

Lorentz Force Equation

The Lorentz force equation is central to our understanding of how charged particles behave in electromagnetic fields. It describes the total force experienced by a charged particle due to both electric and magnetic fields.The Lorentz force equation is given by:\[ \mathbf{F} = q(\mathbf{E} + \mathbf{v} \times \mathbf{B}) \]where:

• \(\mathbf{F}\) is the total force on the particle,
• \(q\) is the charge,
• \(\mathbf{E}\) is the electric field,
• \(\mathbf{v}\) is the velocity of the particle,
• \(\mathbf{B}\) is the magnetic field.

This equation merges the effects of electric and magnetic forces, allowing us to calculate the net force acting on a moving charge. In many situations, such as in the given exercise, the electric field component can be zero, focusing our attention solely on the magnetic influence. Ensuring clarity in such exercises aids in visualizing and calculating the forces at play.