Biot-savart Law And Field Of A Circular Loop
A circular coil of radius 10 cm carries a steady current of 2 A; determine the magnitude of the magnetic field produced at the centre of this single-turn coil.
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About This Question
- Subject
- physics
- Chapter
- magnetic effects of current and magnetism
- Topic
- biot-savart law and field of a circular loop
- Difficulty
- Medium
- Year
- 2025
Solution
Correct Answer:
1.26×10−5 T
According to NCERT Class 12, Chapter 4 (Moving Charges and Magnetism), the Biot-Savart law gives the field contribution of each small current element, and integrating these contributions around a full circular loop of radius R carrying current I yields a field at the centre of B=2Rμ0I. Every element of the loop contributes a field in the same axial direction at the centre, so the contributions add constructively rather than cancelling. This field points along the axis, its direction given by the right-hand rule applied to the sense of the current. Substituting μ0=4π×10−7 T·m/A, I=2 A, and R=0.1 m: B=2(0.1)(4π×10−7)(2)=0.28π×10−7=4π×10−6≈1.26×10−5 T. The value 2.51×10−5 T is double the answer, obtained by forgetting the factor of 2 in the denominator. The value 6.28×10−5 T uses 2π incorrectly in the numerator. The value 1.26×10−6 T is off by a factor of ten from a radius unit error. Plausibility check: fields of order 10−5 T are typical for small laboratory coils carrying a few amperes, comparable to Earth's field, so the answer is reasonable in magnitude.
This medium difficulty physics question is from the chapter magnetic effects of current and magnetism, covering the topic of biot-savart law and field of a circular loop. It appeared in the 2025 exam.
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