Problem 33
Question
An electron and a proton enter a magnetic field perpendicularly. Both have same kinetic energy. Which of the following is true ? (a) Trajectory of electron is less curved (b) Trajectory of proton is less curved (c) Both trajectories are equally curved (d) Both more on straight line path
Step-by-Step Solution
Verified Answer
(b) Trajectory of proton is less curved.
1Step 1: Understanding the Forces
When a charged particle enters a magnetic field perpendicularly, it experiences a force due to the Lorentz force law, which causes it to move in a circular path. The force is given by the equation: \( F = qvB \) where \( q \) is the charge, \( v \) is the velocity, and \( B \) is the magnetic field strength. The force acts as the centripetal force.
2Step 2: Relating Force to Radius
The centripetal force needed for circular motion is given by \( F = \frac{mv^2}{r} \), where \( m \) is the mass, \( v \) is the velocity, and \( r \) is the radius of curvature of the trajectory. By equating the magnetic force to the centripetal force, we have: \( qvB = \frac{mv^2}{r} \). Thus, \( r = \frac{mv}{qB} \).
3Step 3: Using Kinetic Energy
Since both the electron and proton have the same kinetic energy, \( KE = \frac{1}{2}mv^2 \). Therefore, \( v = \sqrt{\frac{2KE}{m}} \). Substituting this into the radius equation, \( r = \frac{m \sqrt{\frac{2KE}{m}}}{qB} = \frac{\sqrt{2mKE}}{qB} \).
4Step 4: Comparing Electron and Proton
The electron and proton both have the same kinetic energy. However, the proton is much more massive than the electron, \( m_p \gg m_e \). Plugging this into the expression for \( r \), we see that \( r_p = \frac{\sqrt{2m_pKE}}{q_pB} \) and \( r_e = \frac{\sqrt{2m_eKE}}{q_eB} \). Because \( m_p \gg m_e \), \( r_p \gg r_e \), meaning the proton's trajectory is less curved.
5Step 5: Conclusion
Given the relation between mass and radius derived, the trajectory of the proton is less curved compared to the electron because of its much larger mass at the same kinetic energy.
Key Concepts
Lorentz ForceCentripetal MotionKinetic Energy
Lorentz Force
When a charged particle, such as an electron or proton, moves through a magnetic field, it experiences a specific force known as the Lorentz Force. This force is essential in electromagnetism and describes the impact of magnetic fields on moving charged particles.
The Lorentz Force is the force that acts on a charged particle in a magnetic field and is directly responsible for the particle's trajectory change. The equation for the Lorentz Force is:
The Lorentz Force is the force that acts on a charged particle in a magnetic field and is directly responsible for the particle's trajectory change. The equation for the Lorentz Force is:
- \( F = qvB \)
- \( F \) is the force acting on the particle,
- \( q \) is the charge of the particle,
- \( v \) is the velocity of the particle,
- \( B \) is the magnetic field strength.
Centripetal Motion
Centripetal Motion refers to the curved path an object follows as it moves around a circle. For a charged particle in a magnetic field, this occurs because the Lorentz Force acts as a centripetal force.
To keep a particle moving in a circle, it must continuously change direction, which requires a centripetal force. This force is what keeps the particle rotating around a central point. It's quantified by:
To keep a particle moving in a circle, it must continuously change direction, which requires a centripetal force. This force is what keeps the particle rotating around a central point. It's quantified by:
- \( F = \frac{mv^2}{r} \)
- \( F \) is the centripetal force,
- \( m \) is the mass of the particle,
- \( v \) is the velocity,
- \( r \) is the radius of curvature.
- \( r = \frac{mv}{qB} \)
Kinetic Energy
Kinetic Energy is the energy a particle possesses as it moves due to its mass and velocity. In this context, it explains why particles like electrons and protons behave differently in a magnetic field.
The Kinetic Energy (KE) of a particle is given by:
By substituting into the radius formula:
The Kinetic Energy (KE) of a particle is given by:
- \( KE = \frac{1}{2}mv^2 \)
By substituting into the radius formula:
- \( v = \sqrt{\frac{2KE}{m}} \)
- \( r = \frac{\sqrt{2mKE}}{qB} \)
Other exercises in this chapter
Problem 31
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