Biot-Savart + Ampere's Law
Magnetic Effects of Current
50
JEE Qs
15%
Hard
80
min
Master the conditions and methods for applying Ampere's Law effectively, as it significantly simplifies B-field calculations for symmetric current distributions compared to Biot-Savart Law's often complex integrations.
🧮 Key Formulas
✅ Key Points for JEE
- 1Biot-Savart Law is fundamental for calculating the magnetic field (B-field) due to any current element, involving vector cross products and integration, always applicable.
- 2Ampere's Law is a powerful simplification for highly symmetric current distributions (like infinite wires, solenoids, toroids) and is analogous to Gauss's Law in electrostatics.
- 3The direction of the B-field is crucial; it's determined by the right-hand thumb rule (for straight wires) or right-hand curl rule (for loops/solenoids), consistent with the direction of the vector cross product in Biot-Savart Law.
- 4When applying Ampere's Law, choose an Amperian loop where the B-field is either constant and parallel to dl_vec, perpendicular to dl_vec, or zero, to simplify the line integral.
- 5The 'I_enc' in Ampere's Law represents the net current piercing through the Amperian loop, considering the direction of current flow according to the right-hand screw rule relative to the loop's orientation.
⚠️ Common Mistakes
- ✕Incorrectly determining the direction of the magnetic field using right-hand rules or making errors in the vector cross product (dl x r).
- ✕Attempting to apply Ampere's Law to problems lacking sufficient symmetry, leading to incorrect or unsolvable integrals.
- ✕Misidentifying the 'I_enc' (net enclosed current) in Ampere's Law, especially with multiple currents or current distributions, or forgetting to account for the direction of current flow.
- ✕Errors in setting up the limits or variables during integration when using Biot-Savart Law for continuous current distributions.
📝 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 Part 1 Ch 4: Moving Charges and Magnetism