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Chapter 27: Problem 23

A particle with initial velocity \(\overrightarrow{\boldsymbol{v}}_{0}=\left(5.85 \times 10^{3} \mathrm{~m} / \mathrm{s}\right) \hat{\jmath}\) enters a region of uniform electric and magnetic fields. The magnetic field in the region is \(\overrightarrow{\boldsymbol{B}}=-(1.35 \mathrm{~T}) \hat{\boldsymbol{k}} .\) 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 (b) \(-0.320 \mathrm{nC}\). You can ignore the weight of the particle.

Short Answer

Expert verified
The magnitude of the electric field in case (a) is calculated to be \(|\overrightarrow{E_a}| = 12.72 \times 10^{3}\) N/C and in case (b) is \(|\overrightarrow{E_b}| = 6.36 \times 10^{3}\) N/C. The direction of the electric field is in the positive y-axis for a positive charge and in the negative y-axis for a negative charge.

Step by step solution

01

Calculate the force due to the magnetic field

According to Fleming's left-hand rule, the magnitude of the force \(\overrightarrow{F}_B\) due to a magnetic field on a moving charged particle is given by \(|\overrightarrow{F}_B| = |q|\) \( |\overrightarrow{v}_0| \) \( |\overrightarrow{B}|\), where \(|\overrightarrow{F}_B|\) is the force due the magnetic field, \(|q|\) is the absolute value of the charge of the particle, \(|\overrightarrow{v}_0|\) is the absolute value of the initial velocity of the particle, and \(|\overrightarrow{B}|\) is the strength of the magnetic field. Given that \(|\overrightarrow{v}_0| = 5.85 \times 10^{3}\) ms鈦宦�, and \(|\overrightarrow{B}|= 1.35\) T, we just need to plug the values of the charge of the particle: (a) \(|q|=0.640\) nC and (b) \(|q| = 0.320\) nC, and convert them to Coulombs by multiplying the given values by \(10^{-9}\). With these we can find \(|\overrightarrow{F}_B|\) for each particle.
02

Calculate the Electric field

Since the particle is to pass through undeflected, the magnitude of the electric field that would balance out the magnetic force is given by \(|\overrightarrow{E}| = |\overrightarrow{F}_B| / |q|\). Substituting the magnetic force and the charge calculated in step 1 for each particle gives us the magnitude of the electric field for each case.
03

Find the direction of the Electric field

We can use the right-hand rule to determine the direction of the electric field required to balance the magnetic force such that the net force is zero. With the thumb pointing in the direction of the initial velocity, the middle finger pointing in the direction of the magnetic field, the forefinger will point in the opposite direction to the charge. The direction denoted by forefinger gives the direction of the electric field \(\overrightarrow{E}\), for both cases when charges are positive or negative.

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Key Concepts

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

Magnetic Force
When a charged particle moves through a magnetic field, it experiences a magnetic force. This force is perpendicular to both the velocity of the particle and the magnetic field itself. The magnitude of the magnetic force \( \overrightarrow{F}_B \) can be calculated using the formula:
  • \( |\overrightarrow{F}_B| = |q| \cdot |\overrightarrow{v}_0| \cdot |\overrightarrow{B}| \)
Here, \(|q|\) is the magnitude of the charge, \(|\overrightarrow{v}_0|\) is the velocity, and \(|\overrightarrow{B}|\) is the magnetic field strength. It鈥檚 crucial to note that this force can change the direction but not the speed of the particle. This concept is key when determining how particles will move through fields.
Electric Field
An electric field \( \overrightarrow{E} \) exerts a force on a charged particle, influencing its motion. For a charged particle to pass through an electromagnetic field region undeflected, the forces due to both electric and magnetic fields must balance. The electric field exerts a force \( \overrightarrow{F}_E \) given by:
  • \( |\overrightarrow{F}_E| = |q| \cdot |\overrightarrow{E}| \)
If these fields are balanced, \( |\overrightarrow{E}| = |\overrightarrow{F}_B| / |q| \). By balancing these forces, we ensure the particle's velocity remains constant, allowing it to pass through without deflection.
Particle Motion
Particle motion in electromagnetic fields involves understanding how forces act on the particle. When a particle enters a region with both electric and magnetic fields, its trajectory depends on the resulting net force.
  • If forces are balanced, the particle will continue in a straight line at constant speed.
  • If not balanced, the particle will either curve or accelerate.
By configuring the fields such that magnetic and electric forces cancel each other out, the particle can move undeflected through the region. Understanding this balance is crucial in many applications like particle accelerators and magnetic resonance imaging.
Fleming's Left-Hand Rule
Fleming's Left-Hand Rule is a simple mnemonic to determine the direction of force in a magnetic field. It states that:
  • Thumb: Represents the direction of the Motion of the particle
  • Forefinger: Direction of the magnetic Field
  • Middle finger: Direction of the force (Force on the particle due to magnetic field)
Using this rule, we can visualize how the magnetic force acts on a charged particle moving through a magnetic field, providing insight into how to adjust fields for desired particle motion.
Right-Hand Rule
The Right-Hand Rule helps find the direction of induced electric fields and forces. To use this rule:
  • Thumb: Indicates velocity of the positive charge.
  • Fingers: Point in the direction of the magnetic field.
  • Palm: Faces the direction of the force applied (especially for induced currents).
For negative charges, the direction will be opposite. This rule assists in ensuring that calculations involving force direction, field orientation, and current flow are consistent in electromagnetic principles.

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Most popular questions from this chapter

The large magnetic fields used in MRI can produce forces on electric currents within the human body. This effect has been proposed as a possible method for imaging "biocurrents" flowing in the body, such as the current that flows in individual nerves. For a magnetic field strength of \(2 \mathrm{~T}\), estimate the magnitude of the maximum force on a 1-mm-long segment of a single cylindrical nerve that has a diameter of \(1.5 \mathrm{~mm} .\) Assume that the entire nerve carries a current due to an applied voltage of \(100 \mathrm{mV}\) (that of a typical action potential). The resistivity of the nerve is \(0.6 \Omega \cdot \mathrm{m}\). (a) \(6 \times 10^{-7} \mathrm{~N} ;\) (b) \(1 \times 10^{-6} \mathrm{~N} ;\) (c) \(3 \times 10^{-4} \mathrm{~N}\) (d) \(0.3 \mathrm{~N}\).

An electromagnet produces a magnetic field of \(0.550 \mathrm{~T}\) in a cylindrical region of radius \(2.50 \mathrm{~cm}\) between its poles. A straight wire carrying a current of 10.8 A passes through the center of this region and is perpendicular to both the axis of the cylindrical region and the magnetic field. What magnitude of force does this field exert on the wire?

A particle with charge \(q\) and mass \(m\) is dropped at time \(t=0\) from rest at its origin in a region of constant magnetic field \(\overrightarrow{\boldsymbol{B}}\) that points horizontally. What happens? To answer, construct a Cartesian coordinate system with the \(y\) -axis pointing downward and the \(z\) -axis pointing in the direction of the magnetic field. At time \(t \geq 0\) the particle has velocity \(\overrightarrow{\boldsymbol{v}}=v_{x} \hat{\imath}+v_{y} \hat{\jmath} .\) The net force \(\overrightarrow{\boldsymbol{F}}=F_{x} \hat{\imath}+F_{y} \hat{\jmath}\) on the particle is the vector sum of its weight and the magnetic force. (a) Using Newton's second law, write equations for \(a_{x}\) and \(a_{y},\) where \(\vec{a}=a_{x} \hat{\imath}+a_{y} \hat{\jmath}\) is the acceleration of the particle. (b) Differentiate the second of these equations with respect to time. Then substitute your expression for \(a_{x}=d v_{x} / d t\) to determine an equation for \(d v_{y}^{2} / d t^{2}\) in terms of \(v_{y}\) (c) This result shows that \(v_{y}\) is a simple harmonic oscillator. Use the initial conditions to determine \(v_{y}(t) .\) Write your answer in terms of the angular frequency \(\omega=q B / m\). Note that \(\left(d v_{y} / d t\right)_{0}=g .\) (d) Substitute your result for \(v_{y}(t)\) into your equation for \(d v_{x} / d t .\) Integrate using the initial conditions to determine \(v_{x}(t) .\) (e) Integrate your expressions for \(v_{x}(t)\) and \(v_{y}(t)\) to determine \(x(t)\) and \(y(t) .\) (f) If \(m=1.00 \mathrm{mg}, q=19.6 \mu \mathrm{C}\) and \(B=10.0 \mathrm{~T}\), what maximum vertical distance does the particle drop before returning upward?

A straight, \(2.5 \mathrm{~m}\) wire carries a typical household current of \(1.5 \mathrm{~A}\) (in one direction) at a location where the earth's magnetic field is 0.55 gauss from south to north. Find the magnitude and direction of the force that our planet's magnetic field exerts on this wire if it is oriented so that the current in it is running (a) from west to east, (b) vertically upward, (c) from north to south. (d) Is the magnetic force ever large enough to cause significant effects under normal household conditions?

A wire \(25.0 \mathrm{~cm}\) long lies along the \(z\) -axis and carries a current of \(7.40 \mathrm{~A}\) in the \(+z\) -direction. The magnetic field is uniform and has components \(B_{x}=-0.242 \mathrm{~T}, B_{y}=-0.985 \mathrm{~T},\) and \(B_{z}=-0.336 \mathrm{~T}\) (a) Find the components of the magnetic force on the wire. (b) What is the magnitude of the net magnetic force on the wire?
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