Force on Current-Carrying Conductor
Magnetic Effects of Current
13
JEE Qs
8%
Hard
60
min
Master the vector cross product and related direction rules (Fleming's Left-Hand Rule/Right-Hand Thumb Rule) as they are critical for correctly solving problems involving force direction.
🧮 Key Formulas
✅ Key Points for JEE
- 1The direction of the magnetic force on a current-carrying conductor is given by Fleming's Left-Hand Rule or the right-hand rule for the cross product (L x B).
- 2For a straight conductor of length L in a uniform magnetic field B, the magnitude of the force is F = BILsinθ, where θ is the angle between the direction of current (L vector) and the magnetic field (B vector).
- 3The net magnetic force on a closed current loop placed in a *uniform* magnetic field is always zero, though a net torque may exist.
- 4Parallel current-carrying conductors attract each other, while anti-parallel current-carrying conductors repel each other.
⚠️ Common Mistakes
- ✕Incorrectly determining the direction of the force due to misapplication of Fleming's Left-Hand Rule or the cross product rule.
- ✕Confusing the direction of length vector 'L' with the physical orientation of the wire, especially in complex geometries or when B is not perpendicular to the wire.
- ✕Using the scalar magnitude formula F = BILsinθ when the vector nature of force (F = I(L x B)) is crucial for determining direction in 3D problems.
📝 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