Force on Moving Charge — Lorentz force, circular motion
Magnetic Effects of Current
13
JEE Qs
8%
Hard
75
min
Master the vector cross product for direction and carefully analyze velocity components (parallel and perpendicular to B) to determine the path and apply circular motion dynamics correctly.
🧮 Key Formulas
✅ Key Points for JEE
- 1Magnetic force is always perpendicular to both the velocity vector (v) and the magnetic field vector (B). Therefore, it does no work and cannot change the kinetic energy or speed of the charged particle, only its direction.
- 2The direction of the magnetic force on a positive charge is given by the right-hand rule (v x B). For a negative charge, the force direction is opposite to v x B.
- 3If the initial velocity of the charge is perpendicular to the magnetic field, the particle follows a circular path. The magnetic force provides the necessary centripetal force.
- 4If the initial velocity of the charge has components both parallel (v_parallel) and perpendicular (v_perpendicular) to the magnetic field, the particle follows a helical path. The parallel component remains unchanged, while the perpendicular component causes circular motion.
- 5If the initial velocity is parallel or anti-parallel to the magnetic field, the magnetic force is zero, and the particle continues to move in a straight line with constant velocity.
⚠️ Common Mistakes
- ✕Incorrectly applying the right-hand rule for the direction of the cross product, especially for negative charges.
- ✕Forgetting that magnetic force does no work and therefore cannot change the speed or kinetic energy of the particle.
- ✕Confusing the role of velocity components: only the component perpendicular to B causes circular motion, while the parallel component causes linear motion along B.
- ✕Using vector notation for magnitudes or scalar notation for vectors, leading to errors in calculation or direction.
📝 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