Moving Charge in Crossed Fields — Velocity selector
Magnetic Effects of Current
13
JEE Qs
8%
Hard
45
min
Always visualize the directions of electric force and magnetic force for the given charge and fields, ensuring they are opposite for undeflected motion.
🧮 Key Formulas
✅ Key Points for JEE
- 1In a velocity selector, the electric field (E) and magnetic field (B) are mutually perpendicular to each other and also to the velocity (v) of the charged particle.
- 2The electric force (F_E = qE) and magnetic force (F_M = qvB) act in opposite directions, balancing each other.
- 3For charged particles to pass undeflected, the net force must be zero, meaning the magnitudes of the electric and magnetic forces must be equal (F_E = F_M).
- 4Only charged particles with a specific velocity v = E/B will pass through the crossed fields undeflected, irrespective of their charge or mass.
- 5This principle is crucial in devices like mass spectrometers and cathode ray tubes to select particles of a particular velocity.
⚠️ Common Mistakes
- ✕Incorrectly determining the direction of the magnetic force using the Right Hand Rule, especially for negative charges.
- ✕Confusing the relative directions of E, B, and v, or failing to ensure they are mutually perpendicular as required for the velocity selector condition.
- ✕Forgetting that the charge 'q' cancels out when deriving the selected velocity v = E/B, or incorrectly carrying it through calculations.
📝 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 1: Electric Charges and Fields
- Class 12 Physics Ch 4: Moving Charges and Magnetism