Cyclotron — Working principle
Magnetic Effects of Current
13
JEE Qs
8%
Hard
60
min
Thoroughly understand the distinct roles of the electric and magnetic fields and the critical resonance condition, as these are frequently tested both conceptually and numerically.
🧮 Key Formulas
✅ Key Points for JEE
- 1The magnetic field provides the necessary centripetal force, causing the charged particle to move in semi-circular paths within the dees.
- 2The oscillating electric field across the gap between the dees accelerates the particle, increasing its kinetic energy and speed, which in turn increases the radius of its path.
- 3For continuous acceleration, the frequency of the oscillating electric field must precisely match the cyclotron frequency (f = qB / (2πm)), which is independent of the particle's speed and path radius (for non-relativistic speeds).
- 4The time taken by the particle to complete a semi-circular path inside a dee is constant, independent of its speed or radius, allowing it to remain in resonance with the accelerating electric field.
- 5Limitations include the inability to accelerate uncharged particles, electrons (due to significant relativistic mass variation at high speeds breaking resonance), and the maximum energy being limited by the size of the dees.
⚠️ Common Mistakes
- ✕Confusing the roles of the electric field (acceleration) and the magnetic field (circular path bending).
- ✕Forgetting the resonance condition and the independence of the cyclotron frequency from particle speed/radius (for non-relativistic speeds).
- ✕Applying cyclotron principles to electrons or particles traveling at relativistic speeds without considering mass variation.
- ✕Incorrectly calculating maximum kinetic energy or the required magnetic field/frequency.
📝 Practice Questions
See allQ43.Match List - I with List - II. List - I List - II (A) Magnetic induction (I) Ampere meter 2 (B) Magnetic intensity (II) Weber (C) Magnetic flux (III) Gauss (D) Magnetic moment (IV) Ampere/meter (1) (A)-(I), (B)-(II), (C)-(III), (D)-(IV) (2) (A)-(III), (B)-(IV), (C)-(I), (D)-(II) (3) (A)-(III), (B)-(II), (C)-(I), (D)-(IV) (4) (A)-(III), (B)-(IV), (C)-(II), (D)-(I)
Q46.The magnetic field inside a 200 turns solenoid of radius 10 cm is 2.9 × 10−4Tesla. If the solenoid carries a current of 0.29 A , then the length of the solenoid is ________ πcm .
Q47.A parallel plate capacitor consisting of two circular plates of radius 10 cm is being charged by a constant current of 0.15 A . If the rate of change of potential difference between the plates is 7 × 108 V/s then the integer value of the distance between the parallel plates is ( Take, ϵ0 = 9 × 10−12 mF , π = 227 ) ________ μm . 2025 (29 Jan Shift 2) JEE Main Previous Year Paper
Q37.If B is magnetic field and μ0 is permeability of free space, then the dimensions of (B/μ0) is (1) ML2 T−2 A−1 (2) MT−2 A−1 (3) L−1 A (4) LT−2 A−1
Q46.A proton is moving undeflected in a region of crossed electric and magnetic fields at a constant speed of 2 × 105 ms−1 . When the electric field is switched off, the proton moves along a circular path of radius 2 cm . The magnitude of electric field is x × 104 N/C. The value of x is _______ Take the mass of the proton = 1.6 × 10−27 kg.
Q50.Two long parallel wires X and Y , separated by a distance of 6 cm , carry currents of 5A and 4A, respectively, in opposite directions as shown in the figure. Magnitude of the resultant magnetic field at point P at a distance 2025 (22 Jan Shift 2) JEE Main Previous Year Paper of 4 cm from wire Y is x × 10−5 T. The value of x is__________ . Take permeability of free space as μ0 = 4π × 10−7SI units.
NCERT Chapters
- Class 12 Physics Ch 4: Moving Charges and Magnetism