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The Earth is showered with particles from space known as muons. They have a charge identical to that of an electron but are many times heavier \(\left(m=1.88 \cdot 10^{-28} \mathrm{~kg}\right)\). Suppose a strong magnetic field is established in a lab \((B=0.500 \mathrm{~T})\) and a muon enters this field with a velocity of \(3.00 \cdot 10^{6} \mathrm{~m} / \mathrm{s}\) at a right angle to the field. What will be the radius of the muon's resulting orbit?

Short Answer

Expert verified
The charge of the muon is identical to that of an electron. Answer: The radius of the muon's orbit is approximately \(7.05 \times 10^{-3}\;\mathrm{m}\).

Step by step solution

01

Identify the given values

We are given the following values: Mass of the muon, \(m = 1.88 \times 10^{-28} \mathrm{~kg}\) Velocity of the muon, \(v = 3.00 \times 10^{6} \mathrm{~m/s}\) Magnetic field strength, \(B = 0.500 \mathrm{~T}\) Charge of the muon, \(q = -1.60 \times 10^{-19} \mathrm{~C}\) (since it's identical to the charge of an electron)
02

Use the orbit radius formula

To calculate the radius of the muon's orbit, we can use the formula: \(r = \frac{mv}{qB}\)
03

Substitute the given values and calculate the radius

Now, we can substitute the given values into the formula: \(r = \frac{(1.88 \times 10^{-28}\text{ kg})(3.00 \times 10^6\text{ m/s})}{(-1.60 \times 10^{-19}\text{ C})(0.500\text{ T})}\)
04

Solve the equation

After substituting the numbers, we can simplify the expression to find the radius: \(r = \frac{5.64 \times 10^{-22}}{-8.00 \times 10^{-20}}\) \(r \approx 7.05 \times 10^{-3}\;\mathrm{m}\) So, the radius of the muon's resulting orbit is approximately \(7.05 \times 10^{-3}\;\mathrm{m}\).

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Key Concepts

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

Charged Particle Motion
The motion of charged particles, such as muons, is deeply rooted in the laws of electromagnetism. Whenever a charged particle moves through a magnetic field, it experiences a force that acts perpendicular to both its velocity and the magnetic field. This force doesn't do any work on the particle - it doesn't increase or decrease the particle's energy - but it can change the direction of the particle's velocity. Consequently, if a particle moves into a magnetic field perpendicularly (as with our muon), it follows a circular path.

This phenomenon holds true for all charged particles in magnetic fields, from electrons in cathode ray tubes to protons in the Large Hadron Collider. Understanding this concept allows us to predict and calculate the path of particles in various scientific and technological applications, exemplified by the muon's orbit in our exercise.
Magnetic Force on Moving Charge
The magnetic force on a moving charge is given by the equation \(\mathbf{F} = q(\mathbf{v} \times \mathbf{B})\), where \(q\) is the charge of the particle, \(\mathbf{v}\) is its velocity vector, and \(\mathbf{B}\) is the magnetic field vector. The force is maximal when the charge moves perpendicular to the magnetic field, leading to a circular path called 'cyclotron motion'.

Significance of the Right Angle

In our exercise, the muon enters the magnetic field at a right angle, meaning the angle between \(\mathbf{v}\) and \(\mathbf{B}\) is 90 degrees. This results in the maximum possible magnetic force and, consequently, a uniform circular motion. The direction of the force is determined by the right-hand rule, a useful mnemonic to understand the vector nature of electromagnetism.
Classical Mechanics in Physics
Classical mechanics is the branch of physics that deals with the motion of objects. It is governed by Newton's laws of motion and is applicable to a wide range of macroscopic scenarios. The principle used to solve our muon problem is part of classical mechanics, even though it involves electromagnetic force.

In this context, we consider two main concepts: inertia, which keeps the muon moving in a straight line at a constant speed, and the centripetal force, required for circular motion, provided by the magnetic force. The balance between these forces determines the muon's path in terms of its radius, highlighting the interconnectedness between classical mechanics and electromagnetism.
Cyclotron Motion
Cyclotron motion refers to the spiral path that a charged particle follows within a magnetic field when not entering perpendicularly. In a uniform magnetic field and absent other forces, a charged particle moving perpendicularly will follow a circular path known as cyclotron motion. Its radius is determined by the particle's speed, mass, charge, and the magnetic field's strength, as illustrated in our exercise.

The concept of cyclotron motion lays the foundation for the cyclotron machine, where charged particles are accelerated and their paths are bent using magnetic fields, demonstrating practical applications of circular motion of charged particles in controlled settings. In the case of our muon, the precise calculation of its orbit showcases these underlying principles in a straightforward example.

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Most popular questions from this chapter

An electron is moving with a constant velocity. When it enters an electric field that is perpendicular to its velocity, the electron will follow a _________ trajectory. When the electron enters a magnetic field that is perpendicular to its velocity, it will follow a ____________ trajectory.

A coil consists of \(120 .\) circular loops of wire of radius \(4.80 \mathrm{~cm} .\) A current of 0.490 A runs through the coil, which is oriented vertically and is free to rotate about a vertical axis (parallel to the \(z\) -axis). It experiences a uniform horizontal magnetic field in the positive \(x\) -direction. When the coil is oriented parallel to the \(x\) -axis, a force of \(1.20 \mathrm{~N}\) applied to the edge of the coil in the positive \(y\) -direction can keep it from rotating. Calculate the strength of the magnetic field.

The work done by the magnetic field on a charged particle in motion in a cyclotron is zero. How, then, can a cyclotron be used as a particle accelerator, and what essential feature of the particle's motion makes it possible?

A straight wire with a constant current running through it is in Earth's magnetic field, at a location where the magnitude is \(0.430 \mathrm{G}\). What is the minimum current that must flow through the wire for a \(10.0-\mathrm{cm}\) length of it to experience a force of \(1.00 \mathrm{~N}\) ?

A \(12.0-\mathrm{V}\) battery is connected to a \(3.00-\Omega\) resistor in a rigid rectangular loop of wire measuring \(3.00 \mathrm{~m}\) by \(1.00 \mathrm{~m}\). As shown in the figure, a length \(\ell=1.00 \mathrm{~m}\) of wire at the end of the loop extends into a \(2.00 \mathrm{~m}\) by \(2.00 \mathrm{~m}\) region with a magnetic field of magnitude \(5.00 \mathrm{~T}\) directed into the page. What is the net force on the loop?

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