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Crossed \(\vec{E}\) and \(\vec{B}\) Fields. A particle with initial velocity \(\vec{\boldsymbol{v}}_{0}=\left(5.85 \times 10^{3} \mathrm{m} / \mathrm{s}\right) \hat{\boldsymbol{J}}\) enters a region of uniform electric and magnetic fields. The magnetic field in the region is \(\vec{\boldsymbol{B}}=\) \(-(1.35 \mathrm{T}) \hat{k}\) . Calculate the magnitude and direction of the electric field in the region if the particle is to pass through undeflected, for a particle of charge (a) \(+0.640 \mathrm{nC}\) and \((\mathrm{b})-0.320 \mathrm{nC.}\) You can ignore the weight of the particle.

Short Answer

Expert verified
For (a): \( 7.8975 \times 10^3 \ \text{V/m} \) in \( \hat{i} \) direction; for (b): \( 7.8975 \times 10^3 \ \text{V/m} \) in \( -\hat{i} \) direction.

Step by step solution

01

Understanding the Condition for No Deflection

For the particle to move undeflected through crossed electric \( \vec{E} \) and magnetic \( \vec{B} \) fields, the electric force \( \vec{F}_e = q \vec{E} \) and the magnetic force \( \vec{F}_m = q( \vec{v} \times \vec{B}) \) must cancel each other out. This yields \( \vec{E} = \vec{v} \times \vec{B} \).
02

Cross Product Calculation

Substitute \( \vec{v} = v_0 \hat{j} \) and \( \vec{B} = -B \hat{k} \) into the cross product formula \( \vec{v} \times \vec{B} \). Here, \( v_0 = 5.85 \times 10^3 \ \text{m/s} \) and \( B = 1.35 \ \text{T} \). The cross product is computed as: \[ \vec{v} \times \vec{B} = (v_0 \hat{j}) \times (-B \hat{k}) = v_0 B \hat{i} \]. This results in \[ \vec{E} = (5.85 \times 10^3 \ \text{m/s})(1.35 \ \text{T}) \hat{i} \].
03

Calculating Electric Field Magnitude

Find the magnitude of \( \vec{E} \) computed in the cross product: \[ |\vec{E}| = (5.85 \times 10^3)(1.35) = 7.8975 \times 10^3 \ \text{V/m} \].
04

Electric Field Direction for Different Charges

Since the cross product \( \vec{v} \times \vec{B} \) is directed along the \( \hat{i} \) direction, \( \vec{E} \) must point in the same direction for a positive charge and opposite for a negative charge to allow the particle to pass undeflected. For (a) \( +0.640 \ \text{nC} \), \( \vec{E} \) points in the \( \hat{i} \) direction. For (b) \( -0.320 \ \text{nC} \), \( \vec{E} \) must point in the \( -\hat{i} \) direction.

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Key Concepts

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

Cross Product in Physics
In physics, the cross product is a mathematical operation that helps us find a vector that is perpendicular to two other vectors. This is particularly useful when dealing with forces in electromagnetic fields. The cross product of two vectors \( \vec{A} \) and \( \vec{B} \) is denoted as \( \vec{A} \times \vec{B} \). It is important to note that the result of a cross product is always a vector.

To perform a cross product, consider two vectors in three-dimensional space represented as \( \vec{A} = A_x \hat{i} + A_y \hat{j} + A_z \hat{k} \) and \( \vec{B} = B_x \hat{i} + B_y \hat{j} + B_z \hat{k} \). The cross product is given by:
  • \( \vec{A} \times \vec{B} = (A_y B_z - A_z B_y) \hat{i} + (A_z B_x - A_x B_z) \hat{j} + (A_x B_y - A_y B_x) \hat{k} \)
This formula calculates a new vector that is orthogonal (perpendicular) to both \( \vec{A} \) and \( \vec{B} \).

In the context of this physics problem, the velocity vector \( \vec{v} \) and the magnetic field vector \( \vec{B} \) are considered. By taking their cross product, we can find the direction of the magnetic force acting on the particle.
Particle Motion in Fields
When a charged particle moves through electric \( \vec{E} \) and magnetic \( \vec{B} \) fields, it experiences forces that can change its path. For the particle in question to remain undeflected, the net force acting on it must be zero.

The forces involved are the electric force \( \vec{F}_e = q \vec{E} \) and the magnetic force \( \vec{F}_m = q(\vec{v} \times \vec{B}) \), where \( q \) is the charge of the particle. For the condition of no deflection, these forces must counterbalance each other:
  • \( \vec{F}_e + \vec{F}_m = 0 \)
This implies that the electric field \( \vec{E} \) must be such that \( \vec{E} = \vec{v} \times \vec{B} \), thereby perfectly negating the effects of the magnetic force on the particle's motion.

Understanding this principle is key for applications involving electromagnetic fields, such as in particle accelerators and other scientific equipment that manipulate charged particle paths.
Electric and Magnetic Forces
Electric and magnetic forces are fundamental in understanding how charged particles interact with fields.
  • Electric Force: This force depends on the electric field \( \vec{E} \) and is given by \( \vec{F}_e = q \vec{E} \). It acts in the direction of the electric field for a positive charge and opposite for a negative charge.
  • Magnetic Force: This is influenced by the velocity of the particle \( \vec{v} \) and the magnetic field \( \vec{B} \). The force is expressed as \( \vec{F}_m = q( \vec{v} \times \vec{B}) \) and acts perpendicular to both the velocity and magnetic field vectors.

The particle's charge also determines the direction of these forces, as illustrated in the exercise. For \( +0.640 \ ext{nC} \), the electric force points in the same direction as the cross product result \( \hat{i} \), while for \( -0.320 \ ext{nC} \), it points in the opposite direction. These principles highlight the delicate balance and interactions between electric and magnetic components when managing particle trajectories.

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Most popular questions from this chapter

When a particle of charge \(q > 0\) moves with a velocity of \(\vec { \boldsymbol { v } } _ { 1 }\) at \(45.0 ^ { \circ }\) from the \(\pm x\) -axis in the \(x y\) -plane, a uniform magnetic field exerts a force \(F _ { 1 }\) along the \(- z\) -axis (Fig. \(P 27.55 ) .\) When the same particle moves with a velocity \(\vec { \boldsymbol { v } } _ { 2 }\) with the same magnitude as \(\vec { \boldsymbol { v } } _ { 1 }\) but along the \(+ z\) -zaxis, a force \(\vec { \boldsymbol { F } } _ { 2 }\) of magnitude \(F _ { 2 }\) is exerted on it along the \(+ x\) -axis. (a) What are the magnitude (in terms of \(q , v _ { 1 } ,\) and \(F _ { 2 }\) ) and direction of the magnetic field? (b) What is the magnitude of \(\vec { F } _ { 1 }\) in terms of \(F _ { 2 } ?\)

(a) What is the speed of a beam of electrons when the simultaneous influence of an electric field of \(1.56 \times 10^{4} \mathrm{V} / \mathrm{m}\) and a magnetic field of \(4.62 \times 10^{-3} \mathrm{T},\) with both fields normal to the beam and to each other, produces no deflection of the electrons? (b) In a diagram, show the relative orientation of the vectors \(\vec{\boldsymbol{v}}, \vec{\boldsymbol{E}},\) and \(\vec{\boldsymbol{B}}\) . (c) When the electric field is removed, what is the radius of the electron orbit? What is the period of the orbit?

A particle of charge \(q > 0\) is moving at speed \(v\) in the \(+ z\) -direction through a region of uniform magnetic field \(\vec { \boldsymbol { B } }\) . The magnetic force on the particle is \(\vec { \boldsymbol { F } } = F _ { 0 } ( 3 \hat { \boldsymbol { \imath } } + 4 \hat { \boldsymbol { J } } ) ,\) 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 \(6 F _ { 0 } / q v ,\) determine as much as you can about the remaining components of \(\vec { B } .\)

A particle with charge 7.80\(\mu \mathrm{C}\) is moving with velocity \(\vec{\boldsymbol{v}}=-\left(3.80 \times 10^{3} \mathrm{m} / \mathrm{s}\right) \hat{\boldsymbol{J}}\) . The magnetic force on the particle is measured to be \(\vec{\boldsymbol{F}}=+\left(7.60 \times 10^{-3} \mathrm{N}\right) \hat{\boldsymbol{\imath}}-\left(5.20 \times 10^{-3} \mathrm{N}\right) \hat{\boldsymbol{k}}\) (a) Calculate all the components of the magnetic field you can from this information. (b) Are there components of the magnetic field that are not determined by the measurement of the force? Explain. (c) Calculate the scalar product \(\vec{B} \cdot \vec{F} .\) What is the angle between \(\vec{B}\) and \(\vec{\boldsymbol{F}} ?\)

A cyclotron is to accelerate protons to an energy of 5.4 MeV. The superconducting electromagnet of the cyclotron produces a 2.9 -T magnetic field perpendicular to the proton orbits. (a) When the protons have achieved a kinetic energy of 2.7\(\mathrm { MeV }\) what is the radius of their circular orbit and what is their angular speed? (b) Repeat part (a) when the protons have achieved their final kinetic energy of 5.4\(\mathrm { MeV }\) .

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