Q.24
An electron is accelerated from rest through a potential difference of 400 V. The electron then enters a uniform magnetic field that is perpendicular to the direction of electrons. The radius of the circular path experienced by the electron is 10 cm. The angular speed of electrons, in radians/sec, is
Given data: Charge of electron = 1.6 × 10-19 C; Mass of electron = 9.1 × 10-31 Kg
Problem Setup
An electron gains kinetic energy from acceleration through 400 V, then moves in a circular path of radius 10 cm (0.1 m) in a perpendicular uniform magnetic field. Use electron charge e = 1.6 × 10⁻¹⁹ C and mass m = 9.1 × 10⁻³¹ kg to find angular speed ω.
Step-by-Step Solution
KE = eV = (1.6 × 10⁻¹⁹)(400) = 6.4 × 10⁻¹⁷ JSpeed v follows from ½mv² = KE, so v = √(2KE/m) = √(2 × 6.4 × 10⁻¹⁷/9.1 × 10⁻³¹) = 1.186 × 10⁷ m/s.
mv²/r = evBB = mv/(er) = (9.1 × 10⁻³¹ × 1.186 × 10⁷)/(1.6 × 10⁻¹⁹ × 0.1) = 6.745 × 10⁻⁴ T
ω = v/r = 1.186 × 10⁷/0.1 = 1.186 × 10⁸ rad/s(or equivalently
ω = eB/m)
Option Analysis
| Option | Value | Status | Reason |
|---|---|---|---|
| (A) | 1.18 × 10⁷ | ❌ Wrong | This is linear speed v, not ω = v/r |
| (B) | 1.18 × 10⁸ | ✅ Correct | Matches calculated ω |
| (C) | 2.18 × 10⁷ | ❌ Incorrect | Possibly from doubling v or proton mass error |
| (D) | 2.18 × 10⁸ | ❌ Wrong | Might arise from √2 error in v or doubled charge |
Key Formulas Summary
v = √(2eV/m) → Speed from potential differencer = mv/(eB) → Cyclotron radiusω = v/r = eB/m → Angular speed (Cyclotron frequency)
Common Exam Mistakes
| Mistake | Result | Why Wrong |
|---|---|---|
Using v as ω |
1.18 × 10⁷ | Forgets ω = v/r |
| Wrong KE calculation | ~0.84 × 10⁷ | Misapplies energy formula |
| Proton mass instead | Lower ω |
Electron-specific values required |
[web:5][code_file:1]
CSIR NET Preparation Tips
- Practice electron magnetic field radius variations: change V, r, or find B
- Always verify units (
rin meters, not cm) - Master Lorentz force
F = evB sinθforθ=90° - Remember
ω = eB/mis independent of velocity!
[web:1][web:3]
🎯 Final Answer
Option (B) 1.18 × 10⁸ rad/s is correct for this CSIR NET physics question.
