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Chapter 28: Problem 74

A particle with charge \(2.0\) C moves through a uniform magnetic field. At one instant the velocity of the particle is \((2.0 \hat{\mathrm{i}}+4.0 \hat{\mathrm{j}}+6.0 \hat{\mathrm{k}}) \mathrm{m} / \mathrm{s}\) and the magnetic force on the particle is \((4.0 \hat{i}-20)+12 \hat{k}) \mathrm{N}\). The \(x\) and \(y\) components of the magnetic field are equal. What is \(\vec{B}\) ?

Short Answer

Expert verified
\( \vec{B} = -3 \hat{i} - 3 \hat{j} - 4 \hat{k} \)

Step by step solution

01

Identify Given Values and Variables

We are given the charge of the particle as \( q = 2.0 \) C, the velocity \( \vec{v} = 2.0 \hat{i} + 4.0 \hat{j} + 6.0 \hat{k} \text{ m/s} \) and the magnetic force \( \vec{F} = 4.0 \hat{i} - 20 \hat{j} + 12 \hat{k} \text{ N} \). We need to find the magnetic field \( \vec{B} \), knowing that the \( x \) and \( y \) components of \( \vec{B} \) are equal.
02

Use the Magnetic Force Formula

The magnetic force on a charged particle is given by the equation \( \vec{F} = q(\vec{v} \times \vec{B}) \). This implies \( \vec{v} \times \vec{B} = \frac{\vec{F}}{q} \). Substitute \( \vec{F} \) and \( q \) to get \( \vec{v} \times \vec{B} = \left(\frac{4.0}{2.0} \right) \hat{i} + \left(\frac{-20}{2.0} \right) \hat{j} + \left(\frac{12}{2.0} \right) \hat{k} \), which simplifies to \( \vec{v} \times \vec{B} = 2\hat{i} - 10\hat{j} + 6\hat{k} \).
03

Set Components of \( \vec{B} \)

Assume the magnetic field is \( \vec{B} = b \hat{i} + b \hat{j} + B_z \hat{k} \) since the x and y components are the same. Then calculate \( \vec{v} \times \vec{B} \).
04

Calculate \( \vec{v} \times \vec{B} \)

Using the cross product formula, \( \vec{v} \times \vec{B} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 2 & 4 & 6 \ b & b & B_z \end{vmatrix} \). \( \vec{v} \times \vec{B} \) resolves to \( (4B_z - 6b)\hat{i} - (2B_z - 6b)\hat{j} + (2b - 4b)\hat{k} \).
05

Equate Components with \( \vec{F} \)

Now equate components from \( \vec{v} \times \vec{B} = 2\hat{i} - 10\hat{j} + 6\hat{k} \) to get: 1. \( 4B_z - 6b = 2 \) 2. \( -2B_z + 6b = -10 \) 3. \( -2b = 6 \)
06

Solve the Equation System

Solve the third equation \( -2b = 6 \) to find \( b = -3 \). Substitute \( b = -3 \) in the remaining equations: 1. \( 4B_z + 18 = 2 \) gives \( B_z = -4 \). Thus, \( \vec{B} = -3 \hat{i} - 3 \hat{j} - 4 \hat{k} \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Charged Particle
A charged particle is any particle with an electric charge, positive or negative. In our example, the particle has a charge of 2.0 coulombs. This determines how the particle interacts with electric and magnetic fields.
  • Charge determines the direction and magnitude of the force experienced in electromagnetic fields.
  • Positive charges move in one direction in a magnetic field while negative charges move in the opposite direction.
In physics, understanding the charge of a particle is crucial because it influences the behavior and interactions of the particle with other particles and fields.
Magnetic Force
Magnetic force is the force exerted by a magnetic field on a moving charged particle. For a particle to experience a magnetic force, it must be moving, since magnetic forces act perpendicular to the velocity of the particle.
  • The magnetic force can be calculated using the equation: \( \vec{F} = q(\vec{v} \times \vec{B}) \).
  • The magnetic force depends on the charge of the particle, its velocity, and the magnetic field.
  • It acts perpendicular to both the velocity of the particle and the magnetic field.
Magnetic force ensures that the motion of charged particles is often curved when within a magnetic field, leading to circular or helical paths depending on other present forces.
Cross Product Formula
To better understand the magnetic force on a charged particle, we use the cross product formula. The cross product, denoted as \( \times \), is a vector operation that takes two vectors and returns a third vector.
  • The vectors must be in a three-dimensional space.
  • It is represented mathematically as: \( \vec{v} \times \vec{B} \).
In the context of magnetic force, this product gives the resulting direction and magnitude of the force acting on the charged particle.In our exercise, the cross product formula helped determine the components of the magnetic field: - The cross product matrix is set up as: \[ \vec{v} \times \vec{B} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ v_x & v_y & v_z \ B_x & B_y & B_z \end{vmatrix} \] leading to a detailed calculation of each component.By solving the resulting system of equations from the cross product, one can find the values of the magnetic field components.

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Most popular questions from this chapter

A proton circulates in a cyclotron, beginning approximately at rest at the center. Whenever it passes through the gap between dees, the electric potential difference between the dees is \(200 \mathrm{~V}\). (a) By how much does its kinetic energy increase with each passage through the gap? (b) What is its kinetic energy as it completes 100 passes through the gap? Let \(r_{100}\) be the radius of the proton's circular path as it completes those 100 passes and enters a dee, and let \(r_{101}\) be its next radius, as it enters a dee the next time. (c) By what percentage does the radius increase when it changes from \(r_{100}\) to \(r_{101} ?\) That is, what is $$ \text { percentage increase }=\frac{r_{101}-r_{100}}{r_{100}} 100 \% ? $$

An electron with kinetic energy \(2.5 \mathrm{keV}\) moving along the positive direction of an \(x\) axis enters a region in which a uniform electric field of magnitude \(10 \mathrm{kV} / \mathrm{m}\) is in the negative direction of the \(y\) axis. A uniform magnetic field \(\vec{B}\) is to be set up to keep the electron moving along the \(x\) axis, and the direction of \(\vec{B}\) is to be chosen to minimize the required magnitude of \(\vec{B}\). In unit-vector notation, what \(\vec{B}\) should be set up?

A beam of electrons whose kinetic energy is \(K\) emerges from a thin-foil "window" at the end of an accelerator tube. A metal plate at distance \(d\) from this window is perpendicular to the direction of the emerging beam (Fig. 28-53). (a) Show that we can prevent the beam from hitting the plate if we apply a uniform magnetic field such that $$ B \geq \sqrt{\frac{2 m K}{e^{2} d^{2}}} $$ in which \(m\) and \(e\) are the electron mass and charge. (b) How should \(\vec{B}\) be oriented?

An electron has an initial velocity of \((12.0 \hat{\mathrm{j}}+15.0 \mathrm{k}) \mathrm{km} / \mathrm{s}\) and a constant acceleration of \(\left(2.00 \times 10^{12} \mathrm{~m} / \mathrm{s}^{2}\right) \hat{\mathrm{i}}\) in a region in which uniform electric and magnetic fields are present. If \(\vec{B}=(400 \mu \mathrm{T}) \hat{\mathrm{i}}\), find the electric field \(\vec{E}\).

A wire of length \(25.0 \mathrm{~cm}\) carrying a current of \(4.51 \mathrm{~mA}\) is to be formed into a circular coil and placed in a uniform magnetic field \(\vec{B}\) of magnitude \(5.71 \mathrm{mT}\). If the torque on the coil from the field is maximized, what are (a) the angle between \(\vec{B}\) and the coil's magnetic dipole moment and (b) the number of turns in the coil? (c) What is the magnitude of that maximum torque?
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