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(II) For a particle of mass \(m\) and charge \(q\) moving in a circular path in a magnetic field \(B,(a)\) show that its kinetic energy is proportional to \(r^{2},\) the square of the radius of curvature of its path, and (b) show that its angular momentum is \(L=q B r^{2},\) about the center of the circle.

Short Answer

Expert verified
Kinetic energy is proportional to \( r^2 \) and angular momentum \( L = q B r^2 \).

Step by step solution

01

Understanding the problem

The exercise asks us to analyze a particle of mass \( m \) and charge \( q \) that is moving in a circular path under the influence of a magnetic field \( B \). We need to show two things: (a) the kinetic energy is proportional to the square of the radius \( r^2 \), and (b) the angular momentum about the center of the circle is \( L = q B r^2 \).
02

Relate the magnetic force to circular motion

For a charged particle moving in a magnetic field, the magnetic force provides the centripetal force necessary for circular motion. This can be expressed as: \( q v B = \frac{m v^2}{r} \), where \( v \) is the velocity of the particle and \( r \) is the radius of the circular path.
03

Solve for velocity

Rearranging the equation from Step 1, we find the velocity: \( v = \frac{q B r}{m} \). This equation shows the relationship between the particle's velocity, its charge, the magnetic field, and the radius of its path.
04

Express kinetic energy

The kinetic energy (KE) of the particle is given by \( KE = \frac{1}{2} m v^2 \). Substitute \( v = \frac{q B r}{m} \) from Step 2 into the kinetic energy formula, resulting in \( KE = \frac{1}{2} m \left(\frac{q B r}{m}\right)^2 \). Simplifying, we get \( KE = \frac{1}{2} \frac{(q B r)^2}{m} \), which shows that \( KE \propto r^2 \).
05

Determine angular momentum

Angular momentum \( L \) for a particle moving in a circle is given by \( L = m v r \). Using the velocity \( v = \frac{q B r}{m} \) from Step 2, substitute it into the angular momentum formula: \( L = m \left(\frac{q B r}{m}\right) r = q B r^2 \). This confirms the required form for angular momentum: \( L = q B r^2 \).

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Key Concepts

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

Kinetic Energy and Radius Relationship
When a charged particle moves through a magnetic field, it tends to follow a circular path due to the magnetic force. A fascinating aspect of this motion is how the kinetic energy of the particle links with the radius of the circle it travels in. Let's delve into this concept.

The kinetic energy (KE) of a particle is defined as the energy it possesses due to its motion, calculated using the formula:
  • \( KE = \frac{1}{2} mv^2 \)
Here, \( m \) represents the mass of the particle, and \( v \) is its velocity.

In a magnetic field, the force exerted on the moving charge acts as the centripetal force which governs circular motion. The unique interplay between this magnetic force and circular motion gives rise to the expression for velocity:
  • \( v = \frac{q Br}{m} \)
where \( q \) is the charge, \( B \) is the magnetic field strength, and \( r \) denotes the radius of the circular path.

Now, substituting this expression for velocity into the kinetic energy equation leads to an intriguing result. The equation for kinetic energy becomes:
  • \( KE = \frac{1}{2} \frac{(q Br)^2}{m} \)
Simplifying further, it clearly shows that \( KE \propto r^2 \). This indicates that kinetic energy increases with the square of the radius. Simply put, a larger radius corresponds to higher kinetic energy.
Centripetal Force in Circular Motion
Charged particles moving in a magnetic field experience a special type of force that keeps them moving in a circular path. This force, known as centripetal force, is essential for understanding the dynamics of charged particles in magnetic fields.

The magnetic force acting on a charged particle can be described as:
  • \( F_{magnetic} = q v B \)
This force is perpendicular to the velocity of the particle and provides the necessary centripetal force to keep the particle in its circular path:
  • \( F_{centripetal} = \frac{m v^2}{r} \)
For circular motion under a magnetic field, these forces must balance, so we equate them:
  • \( q v B = \frac{m v^2}{r} \)
This balance of forces clarifies why the velocity is dependent on the radius, charge, and magnetic field strength. The condition establishes the fact that the centripetal force required for the motion is exactly supplied by the magnetic force.

This ensures the particle's path remains circular, a fascinating phenomenon where nature perfectly balances electric and magnetic interactions with classical mechanics.
Magnetic Force and Charge Interaction
The interaction between a magnetic field and a moving charge is a core concept in electromagnetism. This interaction is pivotal in understanding the motion of charged particles under the influence of a magnetic field.

The magnetic force experienced by a charged particle moving with a velocity \( v \) in a magnetic field \( B \) is given by:
  • \( F = q v B \)
This force is always perpendicular to both the velocity of the particle and the magnetic field direction, giving rise to circular motion when a particle is constrained to move in a plane.

Now, this perpendicular nature of the force means it does no work on the particle. Thus, it cannot change the particle's speed, only its direction. This is why the radius of the path and velocity are correlated. In fact, the larger the radius, the smaller the force needed to change the direction—explaining the kinetic energy and radius relationship discussed earlier.

Understanding this force allows us to comprehend why the energy involved is solely kinetic, and why angular momentum and other properties in circular motion result from this quintessential interaction between magnetic fields and charge.

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Most popular questions from this chapter

A long wire stretches along the \(x\) axis and carries a 3.0-A current to the right \((+x)\). The wire is in a uniform magnetic field \(\overrightarrow{\mathbf{B}}=(0.20 \hat{\mathbf{i}}-0.36 \hat{\mathbf{j}}+0.25 \hat{\mathbf{k}})\) T. Determine the components of the force on the wire per \(\mathrm{cm}\) of length.

(a) What value of magnetic field would make a beam of electrons, traveling to the right at a speed of \(4.8 \times 10^{6} \mathrm{~m} / \mathrm{s}\) go undeflected through a region where there is a uniform electric field of \(8400 \mathrm{~V} / \mathrm{m}\) pointing vertically up? \((b)\) What is the direction of the magnetic field if it is known to be perpendicular to the electric field? \((c)\) What is the frequency of the circular orbit of the electrons if the electric field is turned off?

(II) An electron moves with velocity \(\vec{\mathbf{v}}=(7.0 \hat{\mathbf{i}}-6.0 \hat{\mathbf{j}}) \times 10^{4} \mathrm{m} / \mathrm{s}\) in a magnetic field \(\vec{\mathbf{B}}=(-0.80 \hat{\mathrm{i}}+0.60 \hat{\mathrm{j}}) \mathrm{T}\) . Determine the magnitude and direction of the force on the electron.

A stiff wire \(50.0 \mathrm{~cm}\) long is bent at a right angle in the middle. One section lies along the \(z\) axis and the other is along the line \(y=2 x\) in the \(x y\) plane. A current of \(20.0 \mathrm{~A}\) flows in the wire \(-\) down the \(z\) axis and out the line in the \(x y\) plane. The wire passes through a uniform magnetic field given by \(\overrightarrow{\mathbf{B}}=(0.318 \hat{\mathbf{i}})\) T. Determine the magnitude and direction of the total force on the wire.

Calculate the magnetic force on an airplane which has acquired a net charge of \(1850 \mu \mathrm{C}\) and moves with a speed of \(120 \mathrm{~m} / \mathrm{s}\) perpendicular to the Earth's magnetic field of \(5.0 \times 10^{-5} \mathrm{~T}\).

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