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A particle with charge \(6.40 \times 10^{-19} \mathrm{C}\) travels in a circular orbit with radius 4.68 \(\mathrm{mm}\) due to the force exerted on it by a magnetic field with magnitude 1.65 \(\mathrm{T}\) and perpendicular to the orbit. (a) What is the magnitude of the linear momentum \(\vec{p}\) of the partcle? (b) What is the magnitude of the angular momentum \(\overrightarrow{\boldsymbol{L}}\) of the particle?

Short Answer

Expert verified
(a) The linear momentum \( p \) is \( 4.97 \times 10^{-21} \; kg \cdot m/s \). (b) The angular momentum \( L \) is \( 2.32 \times 10^{-23} \; kg \cdot m^2/s \).

Step by step solution

01

Understanding the Magnetic Force

Magnetic force provides the centripetal force for the particle moving in a circular path. The magnetic force can be expressed as \( F = qvB \), where \( q \) is the charge, \( v \) is the velocity, and \( B \) is the magnetic field strength.
02

Relating Magnetic Force to Centripetal Force

For circular motion, the magnetic force acting as the centripetal force is expressed as \( F = \frac{mv^2}{r} \). We can equate \( qvB = \frac{mv^2}{r} \) to solve for the velocity \( v \) and linear momentum \( p = mv \).
03

Solving for Velocity

From the equation \( qvB = \frac{mv^2}{r} \), we can solve for \( v \) by rearranging it to \( v = \frac{qBr}{m} \). However, we need to express \( m \) in terms of momentum \( p = mv \).
04

Calculating Linear Momentum

Use the rearranged formula \( p = qBr \) for linear momentum. Substitute \( q = 6.40 \times 10^{-19} \; C \), \( B = 1.65 \; T \), and \( r = 0.00468 \; m \) into this equation to find \( p \).
05

Calculating Angular Momentum

Angular momentum \( L \) is given by \( L = mvr \). Since \( p = mv \), we have \( L = pr \). Substitute \( r = 0.00468 \; m \) and the previously calculated \( p \) to find \( L \).

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Key Concepts

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

Linear Momentum
Linear momentum is a fundamental concept in physics and describes the amount of motion a particle has. It is represented by the product of the particle's mass and its velocity. In formula terms, it is denoted as \( p = mv \), where \( p \) is the momentum, \( m \) is the mass, and \( v \) is the velocity of the particle.

- Linear momentum helps us understand how much force is needed to stop an object or how much effect that object would have in a collision.- It's a vector quantity, meaning it has both a magnitude and a direction, aligning with the direction of the velocity.

In the context of the circular motion of a charged particle in a magnetic field, understanding linear momentum becomes crucial. The magnetic force acts continuously perpendicular to the velocity of the particle, causing it to move in a circular path. As a result, the magnetic field doesn't change the speed of the particle but constantly redirects its path. To determine the magnitude of the linear momentum for this case, we use the formula \( p = qBr \), where \( q \) is the charge, \( B \) is the magnetic field strength, and \( r \) is the radius of the circular path. This equation shows how momentum is influenced by magnetic as well as geometric factors, providing a clearer picture of motion dynamics.
Angular Momentum
Angular momentum is the rotational analog of linear momentum. It provides insight into how an object rotates and is given by the formula \( L = mvr \), with \( L \) as the angular momentum, \( m \) as the mass, \( v \) as the tangential velocity, and \( r \) as the radius of the circle.

- It helps us understand rotational dynamics in systems, indicating how quickly and in what manner an object will spin.- Like linear momentum, angular momentum is also a vector, having both magnitude and a direction, which is perpendicular to the plane of rotation.

In the context of a charged particle moving in a circular path due to a magnetic field, angular momentum reveals much about the particle's movement characteristics. The relationship \( L = pr \) emerges due to the fact that linear momentum \( p \) replaces \( mv \) when combined with the radius. This simplification shows that, for circular paths under the influence of central forces like magnetism, angular momentum is contingent upon the radius and linear momentum. It's a perfect way to encapsulate how mass distribution and speed affect rotational behavior in a magnetic field.
Circular Motion
Circular motion is a movement that follows the circumference of a circle and can be found in both everyday phenomena and complex systems. It is a fundamental motion type characterized by a constant distance from a central point, known as the radius.

- In uniform circular motion, the speed remains constant, but the direction of the velocity changes continuously.- The centripetal force is essential to keep the object moving in a circle, preventing it from moving off in a straight line due to inertia.

In our exercise involving a charged particle in a magnetic field, the particle follows a circular path through the action of the magnetic force, which serves as the centripetal force. This means the force due to magnetism is always aimed towards the center of the circle, constantly altering the direction of the velocity of the particle without changing its magnitude. The relationship \( F = qvB \) (for magnetic force), where \( F \) is the force, \( q \) is charge, \( v \) is velocity, and \( B \) is the magnetic field, demonstrates how velocity, charge, and field strength dictate the motion's nature. This snapshot into circular motion under magnetic influence allows a clear understanding of dynamics for particles following magnetic lines in various physics and engineering applications.

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Most popular questions from this chapter

Determining Diet. One method for determining the amount of corn in early Native American diets is the stable isotope ratio analysis (SIRA) technique. As corn photosynthesizes, it concentrates the isotope carbon-13, whereas most other plants concentrate carbon- 12 . Overneliance on corn consumption can then be correlated with certain diseases, because corn lacks the essential amino acid lysine. Archaeologists use a mass spectrometer to separate the \(^{10} \mathrm{C}\) and \(^{13} \mathrm{C}\) isotopes in samples of human remains. Suppose you use a velocity selector to obtain singly ionized (missing one electron) atoms of speed 8.50 \(\mathrm{km} / \mathrm{s}\) , and you want to bend them within a uniform magnetic field in a semicircle of diameter 25.0 \(\mathrm{cm}\) for the 12 \(\mathrm{C}\) . The measured masses of these isotopes are \(1.99 \times 10^{-25} \mathrm{kg}\left(^{12} \mathrm{C}\right)\) and \(2.16 \times 10^{-26} \mathrm{kg}\left(^{13} \mathrm{C}\right) .\) (a) What strength of magnetic field is required? (b) What is the diameter of the \(^{13} \mathrm{C}\) semicircle? (c) What is the separation of the \(^{12}\mathrm{C}\) and \(^{13}\mathrm{C}\) ions at the detector at the end of the semicircle? Is this distance large enough to be easily observed?

A particle with charge \(-5.60 \mathrm{nC}\) is moving in a uniform magnetic field \(\overrightarrow{\boldsymbol{B}}=-(1.25 \mathrm{T}) \hat{\boldsymbol{k}} .\) The magnetic force on the particle is measured to be \(\overrightarrow{\boldsymbol{F}}=-\left(3.40 \times 10^{-7} \mathrm{N}\right) \hat{\boldsymbol{i}}+\left(7.40 \times 10^{-7} \mathrm{N}\right) \hat{\boldsymbol{j}}\) (a) Calculate all the components of the velocity of the particle that you can from this information. (b) Are there components of the velocity that are not determined by the measurement of the force? Explain. (c) Calculate the scalar product \(\overrightarrow{\boldsymbol{F}} \cdot \overrightarrow{\boldsymbol{F}}\) . What is the angle between \(\overrightarrow{\boldsymbol{v}}\) and \(\overrightarrow{\boldsymbol{F}} ?\)

A proton \(\left(q=1.60 \times 10^{-19} \mathrm{C}, m=1.67 \times 10^{-27} \mathrm{kg}\right)\) moves in a uniform magnetic field \(\overrightarrow{\boldsymbol{B}}=(0.500 \mathrm{T}) \hat{\boldsymbol{i}} .\) At \(t=0\) the proton has velocity components \(v_{x}=1.50 \times 10^{5} \mathrm{m} / \mathrm{s}, v_{y}=0,\) and \(v_{z}=2.00 \times 10^{5} \mathrm{m} / \mathrm{s}(\text { see Example } 27.4) .\) (a) What are the magnitude and direction of the magnetic force acting on the proton? In addition to the magnetic field there is a uniform electric field in the \(+x\) -direction, \(\vec{E}=\left(+2.00 \times 10^{4} \mathrm{V} / \mathrm{m}\right) \hat{\imath}\) (b) Will the proton have a component of acceleration in the direction of the electric field?(c) Describe the path of the proton. Does the electric field affect the radius of the helix? Explain. (d) At \(t=T / 2\) , where \(T\) is the period of the circular motion of the proton, what is the \(x\) -component of the displacement of the proton from its position at \(t=0 ?\)

An Electromagnetic Rail Gun. A conducting bar with mass \(m\) and length \(L\) slides over horizontal rails that are connected to a voltage source. The voltage source maintains a constant current \(I\) in the rails and bar, and a constant, uniform, vertical magnetic field \(\overrightarrow{\boldsymbol{B}}\) fills the region between the rails (Fig. 27.63\()\) . (a) Find the magnitude and direction of the net force on the conducting bar Ignore friction, air resistance, and electrical resistance. (b) If the bar has mass \(m\) , find the distance \(d\) that the bar must move along the rails from rest to attain speed \(v\) . (c) It has been suggested that rail guns based on this principle could accelerate payloads into earth orbit or beyond. Find the distance the bar must travel along the rails if it is to reach the escape speed for the earth \((11.2 \mathrm{km} / \mathrm{s}) .\) Let \(B=0.50 \mathrm{T}, \quad I=2.0 \times 10^{3} \mathrm{A}\) \(m=25 \mathrm{kg},\) and \(L=50 \mathrm{cm} .\) For simplicity assume the net force on the object is equal to the magnetic force, as in parts \((a)\) and \((b)\) , even though gravity plays an important role in an actual launch in space.

An electron in the beam of a TV picture tube is accelerated by a potential difference of 2.00 \(\mathrm{kV}\) . Then it passes through a region of transverse magnetic field, where it moves in a circular arc with radius 0.180 \(\mathrm{m}\) . What is the magnitude of the field?

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