Problem 58 - University Physics with Modern Physics

Question

A particle of charge \(q\) > 0 is moving at speed v in the \(+z\)-direction through a region of uniform magnetic field \(\overrightarrow{B}\). The magnetic force on the particle is \(\overrightarrow{F} =\) \(F_0\)(3\(\hat{\imath}\) + 4 \(\hat{\jmath}\)), where \(F_0\) is a positive constant. (a) Determine the components \(B_x\), \(B_y\), and \(B_z\), or at least as many of the three components as is possible from the information given. (b) If it is given in addition that the magnetic field has magnitude 6\(F_0/qv\), determine as much as you can about the remaining components of \(\overrightarrow{B}\).

Step-by-Step Solution

Answer

Components are: \(B_x = \frac{4F_0}{qv}, B_y = -\frac{3F_0}{qv}, B_z = \pm \frac{\sqrt{11}F_0}{qv}\).

Step 1: Understand the Magnetic Force Formula

The magnetic force acting on a moving charged particle is given by the equation \( \overrightarrow{F} = q \cdot (\overrightarrow{v} \times \overrightarrow{B}) \). The cross product \((\overrightarrow{v} \times \overrightarrow{B})\) leads to components in the \(xy\)-plane when the velocity \(\overrightarrow{v}\) is in the \(z\)-direction.

Step 2: Express the Velocity Vector and Force Components

The velocity vector here is \( \overrightarrow{v} = v \hat{k} \), indicating movement in the \(+z\)-direction. The force vector is given as \( \overrightarrow{F} = F_0 (3\hat{\imath} + 4\hat{\jmath}) \). This implies the magnetic field's effect creates a force in the \(xy\)-plane.

Step 3: Cross Product with Velocity in the Z-Direction

Compute the cross-product \( \overrightarrow{v} \times \overrightarrow{B} \). Here, \( \overrightarrow{v} \cdot (v \hat{k}) = 0 \). The result of the cross-product will be in the \(xy\)-plane, meaning contributions from \(B_x\) and \(B_y\) determine \(3\hat{\imath} + 4\hat{\jmath}\).

Step 4: Set Equations for Force Components

Using \( \overrightarrow{F} = q \cdot (v \hat{k} \times (B_x \hat{\imath} + B_y \hat{\jmath} + B_z \hat{k}))\), determine the conditions by setting: \( 3F_0 = qv(-B_y) \) and \( 4F_0 = qv(B_x) \). This yields \(B_x = \frac{4F_0}{qv}\) and \(B_y = -\frac{3F_0}{qv}\).

Step 5: Find the Magnitude Constraint from Given Information

The magnetic field magnitude is \( |\overrightarrow{B}| = \sqrt{(B_x)^2 + (B_y)^2 + (B_z)^2} = \frac{6F_0}{qv}\). Use \(B_x\) and \(B_y\) to compute \(B_z\): \((\frac{4F_0}{qv})^2 + (-\frac{3F_0}{qv})^2 + (B_z)^2 = (\frac{6F_0}{qv})^2\).

Step 6: Solve for the Unknown Component

Simplify to: \( \frac{16F_0^2}{q^2v^2} + \frac{9F_0^2}{q^2v^2} + (B_z)^2 = \frac{36F_0^2}{q^2v^2}\). Thus, \( (B_z)^2 = \frac{36F_0^2}{q^2v^2} - \frac{25F_0^2}{q^2v^2} = \frac{11F_0^2}{q^2v^2} \), yielding \( B_z = \pm \frac{\sqrt{11}F_0}{qv} \).

Key Concepts

Charged Particle Motion

When a charged particle moves through a magnetic field, it experiences a force known as the magnetic force. This force affects the path of the particle based on its charge, speed, and the magnetic field's characteristics. In this exercise, the charged particle with charge \( q \) moves in the positive \( z \)-direction at a speed \( v \). Because it's moving through a magnetic field \( \overrightarrow{B} \), the particle is subject to a magnetic force \( \overrightarrow{F} \).

However, the actual direction of the force is dependent on the orientation of the magnetic field. The magnetic force is always perpendicular to both the velocity of the particle and the magnetic field, following the right-hand rule. In this specific scenario, the force is exerted in the \( xy \)-plane, which means that as the charged particle moves in the \( +z \)-direction, the magnetic field exerts a force that is felt only in the plane perpendicular to its direction of motion. The force's effect will make the particle spiral or curve within this plane, rather than continue in a straight line.

When dealing with problems about charged particle motion, it's crucial to grasp that these particles do not simply follow straight paths; their courses are often altered based on interactions with magnetic fields. Understanding these dynamics is essential in fields like electromagnetism and even in technologies like cyclotrons used in particle physics.

Magnetic Field Components

The magnetic field \( \overrightarrow{B} \) exerting a force on the charged particle can be broken down into its components: \( B_x \), \( B_y \), and \( B_z \). These components describe the field's influence in the \( x \)-, \( y \)-, and \( z \)-directions, respectively.

From the given exercise, we know that the force results entirely in the \( xy \)-plane, implying that the \( z \)-component of the velocity or the force itself doesn't directly interact with any force. Therefore, only \( B_x \) and \( B_y \) contribute to the resultant force \( \overrightarrow{F} \). The solutions indicated that \( B_x = \frac{4F_0}{qv} \) and \( B_y = -\frac{3F_0}{qv} \). These components essentially represent the magnetic field strength in relation to the force and velocity components.

The final piece of the puzzle is the \( B_z \) component, which we understand to not directly result in force due to the velocity's direction along \( z \). However, by using the given magnitude of the magnetic field \( |\overrightarrow{B}| = \frac{6F_0}{qv} \) and the expression \( |\overrightarrow{B}| = \sqrt{(B_x)^2 + (B_y)^2 + (B_z)^2} \), it is possible to solve for \( B_z \) and find that \( B_z = \pm \frac{\sqrt{11}F_0}{qv} \). Understanding these components is crucial when working with magnetic forces, as it determines how a particle behaves under the influence of such fields.

Cross Product in Physics

The cross product is a fundamental concept in physics used to calculate vectors perpendicular to the plane formed by two other vectors. It's specifically significant in determining magnetic force directions. The formula for the magnetic force on a moving charge is \( \overrightarrow{F} = q(\overrightarrow{v} \times \overrightarrow{B}) \), where \( \overrightarrow{v} \) is the velocity vector and \( \overrightarrow{B} \) is the magnetic field vector.

For the cross product \( \overrightarrow{v} \times \overrightarrow{B} \), the result is a vector orthogonal to both \( \overrightarrow{v} \) and \( \overrightarrow{B} \). In this exercise, the velocity vector is \( v \hat{k} \), implying movement in the \( +z \)-direction, while the force produced is in the \( xy \)-plane, showing that \( \overrightarrow{B} \) must necessarily have \( x \) and \( y \) components to result in this plane. These dynamics are dictated by the properties of the cross product.

This mathematical concept highlights that the product produces maximum force when the vectors are perpendicular and no force when they are parallel. As such, learning how to compute and use cross products is vital when dealing with vector quantities like magnetic forces.