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Chapter 27: Problem 51

A charged particle moving in a uniform magnetic field and losses \(4 \%\) of its \(\mathrm{KE}\). The radius of curvature of its path changes by: (a) \(2 \%\) (b) \(4 \%\) (c) \(10 \%\) (d) none of these

Short Answer

Expert verified
The radius changes by 2%.

Step by step solution

01

Understand the Relationship

The kinetic energy (KE) of a charged particle moving in a magnetic field is given by \( KE = \frac{1}{2} mv^2 \), where \( m \) is the mass and \( v \) is the velocity of the particle. The radius of curvature \( r \) is given by \( r = \frac{mv}{qB} \), where \( q \) is the charge and \( B \) is the magnetic field strength.
02

Relate Radius of Curvature to Velocity

From the radius formula \( r = \frac{mv}{qB} \), we know that the radius depends linearly on the velocity \( v \). Therefore, any percentage change in velocity leads to the same percentage change in the radius of curvature \( r \).
03

Calculate Percentage Change in Velocity

Since the particle loses 4% of its kinetic energy, we know that \( \Delta KE = -0.04KE \). The new kinetic energy is \( KE' = 0.96KE \).In terms of velocity, with \( KE = \frac{1}{2} mv^2 \) and \( KE' = \frac{1}{2} m(v')^2 \), we have:\[ 0.96mv^2 = m(v')^2 \]Solving the equation gives:\[ v' = v \sqrt{0.96} \]
04

Determine Percentage Reduction in Velocity

Calculate the approximate percentage change in velocity:\( v' = v \sqrt{0.96} \approx v \times 0.98 \)This calculates the new velocity after losing 4% of KE. Hence, the velocity decreases by approximately 2%.
05

Conclusion of Percentage Change in Radius

As inferred from the relational dependency of radius on velocity, a 2% reduction in velocity results in a 2% reduction in the radius of curvature. Therefore, the radius decreases by 2%.

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Key Concepts

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

Kinetic Energy
Kinetic energy is the energy an object possesses due to its motion. For a charged particle moving in a magnetic field, kinetic energy can be expressed using the formula \( KE = \frac{1}{2} mv^2 \). Here, \( m \) is the mass of the particle and \( v \) is its velocity. This energy is crucial in determining how a particle behaves within a magnetic field.
When the particle in the given problem loses 4% of its kinetic energy, this indicates a reduction in its ability to move, which directly affects its velocity.
Understanding how kinetic energy changes help us predict how other properties like velocity and trajectory are altered.
  • The change in kinetic energy provides information on the relative speed of the charged particle.
  • Even a small percentage change can impact other aspects of a charged particle's motion.
Charged Particle
Charged particles such as electrons and protons are influenced by magnetic fields due to their electric charge. When these particles traverse a magnetic field, they experience a force that changes their direction while maintaining speed, unless acted upon by another force.

In our exercise, the charge of the particle, denoted by \( q \), plays a crucial role in determining the path's radius of curvature. It modifies the path of the particle, causing it to take on a circular trajectory whose curvature depends on its charge, velocity, and the strength of the magnetic field.
  • A charged particle's path is affected by its interaction with the magnetic field.
  • The execution of circular motion in the magnetic field relies on the balance between the centripetal force and the magnetic force, both of which are determined by this charge.
Radius of Curvature
The radius of curvature describes the circular path a charged particle follows in a magnetic field. It is expressed through the formula \( r = \frac{mv}{qB} \), where \( m \) is the mass, \( v \) is the velocity, \( q \) is the charge, and \( B \) is the magnetic field strength.
Any change in the velocity due to variations in kinetic energy will alter the radius. In this exercise, as the kinetic energy decreases by 4%, we can deduce a corresponding change in velocity, affecting the radius as well.

Important points to consider:
  • The radius of curvature is directly proportional to the velocity of the particle.
  • A decrease in kinetic energy leads to a decrease in velocity, thereby reducing the radius of curvature.
  • Understanding this relationship helps explain how the path of a charged particle is altered in different conditions.
Velocity Change
The velocity of a charged particle is an integral aspect of its motion in a magnetic field. Velocity change is often a consequence of energy change. As noted in the exercise, the velocity changes in response to a reduction in kinetic energy.
With a 4% decrease in kinetic energy, we utilized the equation \( v' = v \sqrt{0.96} \) to determine the new velocity. Calculations showed that the new velocity is approximately 98% of the original velocity.
This implies the velocity decreased by about 2%.
  • This velocity change results in a proportional change in the radius of curvature.
  • Understanding how velocity changes informs us about alterations in motion and trajectory.
  • The equation \( v' = v \sqrt{0.96} \) is crucial for translating kinetic energy changes to velocity changes.

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Most popular questions from this chapter

A non-relativistic charged particle of charge \(q\) and mass \(m\) originates at a point at origin of coordinate system. A magnetic field strength \(B\) is directed along \(x\) -axis. The charged particle moves with velocity \(v\) at an angle \(\alpha\) to the \(x\) -axis. A screen is oriented at right angles to the axis and is situated at a distance \(x_{0}\) from the origin of coordinate system. The coordinates of point \(P\) on the screen at which the charged particle strikes: (a) \(x_{0}, \frac{m v \sin \alpha}{q B} \sin \frac{q B}{m} t, \frac{n v \sin \alpha}{q B}\left[1-\frac{\cos q B t}{m}\right]\) (b) \(x_{0}, \frac{m v \sin \alpha}{q B} \sin \frac{q B t}{m}, \frac{m v \sin \alpha}{q B}\left[\cos \frac{q B}{m} t-1\right]\) (c) \(x_{0}, \frac{m \sin \alpha}{q B}\left[\cos \frac{q B}{m} t-1\right] \frac{m v \sin \alpha}{q B} \sin \frac{q B}{m} t\) (d) none of the above

The restoring couple in the moving coil galvanometer is because of : (a) magnetic field (b) material of the coil (c) twist produced in the suspension (d) current in the coil

An electron accelerated through a potential difference \(V\) enters into a uniform transverse magnetic field and experiences a force \(F\). If the accelerating potential is increased to \(2 V\), the electron in the same magnetic field will experience a force: (a) \(\bar{F}\) (b) \(\frac{F}{2}\) (c) \(\sqrt{2} F\) (d) \(2 F\)

A long copper wire carries current in the east direction. The electrons are moving with a drift velocity \(\overrightarrow{\mathbf{u}}\). An. observer now moves with the velocity \(\overrightarrow{\mathbf{u}}\). In the frame of this observer: (a) electric field is present (b) magnetic field due to wire is zero (c) only magnetic field is present (d) none of the above

A charged particle moves in a magnetic field in a plane perpendicular to the magnetic field. The orbital magnetic moment of circulating charge is directed: (a) parallel to magnetic field (b) against the magnetic field (c) perpendicular to magnetic field (d) none of the above
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