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PhysicsMediumClass 12

Biot-Savart Law — Field due to straight wire, circular loop

Magnetic Effects of Current

13

JEE Qs

8%

Hard

75

min

Master the vector cross product and integration techniques, as they are essential for setting up and solving problems involving various current geometries based on Biot-Savart Law.

🧮 Key Formulas

dB = (μ₀ / 4π) * (I * (dl x r_vector) / |r_vector|³)
|dB| = (μ₀ / 4π) * (I * dl * sinθ / r²)
B_straight_wire = (μ₀I / 4πd) * (sinφ₁ + sinφ₂)
B_infinite_straight_wire = μ₀I / 2πd
B_semi_infinite_straight_wire = μ₀I / 4πd
B_center_circular_loop = μ₀I / 2R
B_axis_circular_loop = (μ₀IR² / 2(R² + x²)^(3/2))

✅ Key Points for JEE

  • 1Biot-Savart Law is the fundamental law for calculating the magnetic field produced by current elements (Idl), acting as the magnetic analogue to Coulomb's law for electrostatics.
  • 2The magnetic field (dB) produced by a current element (Idl) is always perpendicular to both Idl and the position vector (r) from the element to the point of observation. Its direction is given by the Right Hand Thumb Rule (or cross product rule).
  • 3For continuous current distributions (straight wires, loops), the total magnetic field (B) is found by integrating the differential field (dB) over the entire current path. This often requires careful setup of integration variables and limits.
  • 4Symmetry plays a crucial role in simplifying calculations; often, components of the magnetic field perpendicular to an axis or plane cancel out.
  • 5Be mindful of the angle definitions in formulas: for the straight wire formula B = (μ₀I / 4πd)(sinφ₁ + sinφ₂), φ₁ and φ₂ are the angles made by the lines joining the observation point to the ends of the wire with the perpendicular line from the observation point to the wire.

⚠️ Common Mistakes

  • Incorrectly applying the Right Hand Thumb Rule for determining the direction of the magnetic field, especially in 3D scenarios or when summing fields from multiple elements.
  • Confusing the angles used in the straight wire formula (φ₁ and φ₂) with angles made by the observation point with the ends of the wire directly, rather than with the perpendicular.
  • Errors in setting up the integral for continuous current distributions, particularly in choosing the differential element (dl), the position vector (r), and the limits of integration.
  • Ignoring the vector nature of the magnetic field and attempting to simply sum magnitudes when the fields from different elements are not in the same direction.

📝 Practice Questions

See all

Q43.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)

2025·MCQEasy

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 .

2025·NumericalMedium

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

2025·NumericalMedium

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

2025·ConceptualEasy

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.

2025·NumericalMedium

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.

2025·NumericalMedium

NCERT Chapters

  • Class 12 Physics Part I Ch 4: Moving Charges and Magnetism