Physics · Moving Charges And Magnetism · NEET
It is 2R. The formula is B = μ₀I/2R. Students confuse it with the straight-wire formula B = μ₀I/2πR, which does have π. Trick: the circular loop already 'used up' the geometry, so its centre field is cleaner with just 2R. If you ever see 2πR for a loop's centre, it is wrong.
In Biot-Savart, dB = (μ₀/4π)(I dl sinθ)/r². For every small element of the loop, the element dl is always perpendicular to the line joining it to the centre, so θ = 90° and sinθ = 1. Also every element is the same distance r = R from the centre. Both simplifications remove the angle term, giving a clean B = μ₀I/2R.
A long straight wire gives B = μ₀I/2πR (has π, field circles the wire). A circular loop's centre gives B = μ₀I/2R (no π, field points along the axis through the centre). Same μ₀I but the loop is stronger at equal R because its whole length wraps around one point.
Yes. N identical turns of radius R stacked at the same place carry current I each, and their fields add: B = μ₀NI/2R. A 100-turn coil gives 100 times the field of a single turn. This is why coils, not single loops, are used to make strong fields.
Along the axis, perpendicular to the plane of the loop. Curl the right hand's fingers along the current direction; the extended thumb points the way B goes at the centre. One face of the loop acts like a North pole (field coming out), the other like South.
A tightly wound 100 turns coil of radius 10 cm carries a current of 7 A. The magnitude of the magnetic field at the centre of the coil is (Take μ₀ = 4π × 10⁻⁷ SI units):
A 100-turn closely wound circular coil of radius 5 cm has a magnetic field of 3.14 × 10⁻³ T at its centre. The current flowing through the coil, and the magnitude of the magnetic moment of this coil are, respectively: (Take μ₀ = 4π × 10⁻⁷ T m/A)
Try the real previous-year questions from this chapter — each with the answer and a full solution.
For a single turn, B = μ₀I/2R. For a coil of N turns of the same radius, B = μ₀NI/2R, where μ₀ = 4π×10⁻⁷ T m/A, I is current in amperes and R is radius in metres.
Take dB = (μ₀/4π)(I dl sinθ)/r². At the centre every element has θ = 90° (sinθ = 1) and r = R. So dB = (μ₀/4π)(I dl)/R². Integrating dl around the loop gives 2πR, so B = (μ₀I/4πR²)(2πR) = μ₀I/2R.
It is along the axis of the loop, perpendicular to its plane. Use the right-hand rule: curl fingers along the current, the thumb shows the field direction at the centre.
Yes, inversely. B = μ₀I/2R, so a smaller radius gives a stronger field at the centre for the same current. Doubling R halves the central field.
The loop formula μ₀I/2R has no π in the denominator, while the straight wire has μ₀I/2πR. Because every part of the loop wraps around the centre, all contributions add up in the same direction, giving a bigger field.
