JEE Physics Magnetism Force on Moving Charge Cyclotron Velocity Selector

What This Is and Why It Matters for JEE

Force on Moving Charge: Cyclotron, Velocity Selector is a fundamental concept in Magnetism. It appears in 2-3 questions every year, making it a moderate difficulty topic. This topic is crucial for JEE Advanced, where it's often tested in more complex forms.

Prerequisites

• Electric Charges and Fields (basic understanding of electric charges, fields, and forces)
• Magnetic Fields (familiarity with magnetic field lines, direction, and strength)
• Motion in a Magnetic Field (basic understanding of force on moving charges in a magnetic field)

Core Concepts (Exam-Focused)

• Force on Moving Charge: F = qvB sin(θ), where F is the force, q is the charge, v is the velocity, B is the magnetic field strength, and θ is the angle between v and B.
• Cyclotron: A device that accelerates charged particles in a circular path using a magnetic field and electric field.
• Velocity Selector: A device that filters charged particles based on their velocity using a magnetic field.
• Key Conditions: The force on a moving charge is maximum when θ = 90°.

Step-by-Step Problem-Solving Strategy

1. Identify the given information: charge, velocity, magnetic field strength, and angle.
2. Check if the force is maximum or minimum. If maximum, θ = 90°.
3. Set up the equation: F = qvB sin(θ).
4. Check for multiple cases or special conditions: e.g., θ = 0° or θ = 180°.
5. ⚠️ Avoid assuming θ = 90° without checking the given information.

Important Graphs / Diagrams

• Magnetic Field Lines: Examiners test the direction and strength of magnetic field lines.
• Force on Moving Charge Graph: A graph showing the force on a moving charge as a function of angle.

Typical JEE Question Patterns

• Find minimum value of...: Identify the minimum value of force on a moving charge.
• Compare time periods...: Compare the time periods of two cyclotrons or velocity selectors.
• Determine the velocity...: Determine the velocity of a charged particle in a magnetic field.

Common Mistakes & Exam Traps

• The mistake: Assuming θ = 90° without checking the given information.
• Why it happens: Misreading or misunderstanding the problem.
• How to avoid it: Carefully read the problem and check the given information.

• Exam board insight: This mistake is penalized in JEE Advanced.

• The mistake: Not considering multiple cases or special conditions.

• Why it happens: Rushing or not reading the problem carefully.
• How to avoid it: Take your time and read the problem carefully.
• Exam board insight: This mistake is penalized in JEE Main.

Time-Saving Shortcuts

• Use the equation F = qvB sin(θ) to quickly calculate the force on a moving charge.
• Check if the force is maximum or minimum by looking at the angle θ.

Practice MCQs (Exam-Style)

Question 1: A charged particle moves with a velocity of 2 × 10^6 m/s in a magnetic field of 0.5 T. What is the force on the particle if the angle between the velocity and magnetic field is 60°?

A) 2 × 10^-6 N
B) 5 × 10^-6 N
C) 8 × 10^-6 N
D) 10 × 10^-6 N

Answer: B) 5 × 10^-6 N
Solution: Use the equation F = qvB sin(θ). Since the angle is 60°, sin(60°) = 0.866. Plug in the values: F = (1.6 × 10^-19 C) × (2 × 10^6 m/s) × (0.5 T) × 0.866 = 5 × 10^-6 N.
Common Wrong Answer: Option D) 10 × 10^-6 N is tempting because it's a multiple of 10^-6. However, the correct answer is 5 × 10^-6 N.

Question 2: A cyclotron accelerates charged particles in a circular path with a radius of 10 cm. What is the time period of the cyclotron if the magnetic field strength is 1 T?

A) 2 × 10^-3 s
B) 5 × 10^-3 s
C) 10 × 10^-3 s
D) 20 × 10^-3 s

Answer: A) 2 × 10^-3 s
Solution: Use the equation T = 2πr / v. Since the particle moves in a circular path, v = 2πr / T. Rearrange the equation to get T = 2πr / v. Plug in the values: T = 2π × (0.1 m) / (2 × 10^6 m/s) = 2 × 10^-3 s.
Common Wrong Answer: Option C) 10 × 10^-3 s is tempting because it's a multiple of 10^-3. However, the correct answer is 2 × 10^-3 s.

Question 3: A velocity selector filters charged particles based on their velocity. What is the velocity of a charged particle that passes through the selector if the magnetic field strength is 0.2 T and the electric field strength is 10^3 V/m?

A) 5 × 10^6 m/s
B) 10 × 10^6 m/s
C) 20 × 10^6 m/s
D) 50 × 10^6 m/s

Answer: B) 10 × 10^6 m/s
Solution: Use the equation v = E / B. Plug in the values: v = (10^3 V/m) / (0.2 T) = 10 × 10^6 m/s.
Common Wrong Answer: Option A) 5 × 10^6 m/s is tempting because it's a multiple of 5 × 10^6. However, the correct answer is 10 × 10^6 m/s.

Quick Revision Card (60-Second Summary)

• F = qvB sin(θ): Force on a moving charge
• Cyclotron: Accelerates charged particles in a circular path
• Velocity Selector: Filters charged particles based on their velocity
• θ = 90°: Maximum force on a moving charge
• T = 2πr / v: Time period of a cyclotron
• v = E / B: Velocity of a charged particle in a velocity selector

If You Get Stuck in Exam

• Write partial marks: Even if unsure, write what you know and get partial marks.
• Eliminate distractors: Eliminate obviously incorrect options.
• Skip and return: Skip the question and return to it later if you're unsure.

Related JEE Topics

• Electric Charges and Fields: Understanding electric charges, fields, and forces is crucial for this topic.
• Magnetic Fields: Familiarity with magnetic field lines, direction, and strength is essential for this topic.
• Motion in a Magnetic Field: Understanding the force on moving charges in a magnetic field is critical for this topic.