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Using Ampère's law, the magnetic field at a perpendicular distance r from a long straight wire carrying current I is:
Aμ₀I / (2πr)
Bμ₀I / (2r)
Cμ₀I / (4πr)
Dμ₀I r / (2π)
Answer & Solution
Correct answer: A. μ₀I / (2πr)
1. Choose a circular amperian loop of radius $r$ around the wire; B is tangential and constant on it.
2. Then $\oint \mathbf{B}\cdot d\mathbf{l} = B(2\pi r)$.
3. Ampère's law: $B(2\pi r) = \mu_0 I$.
4. Solve: $B = \dfrac{\mu_0 I}{2\pi r}$ (the $1/2R$ form belongs to a loop centre, not a straight wire).
_Source: NCERT Class 12 Physics Ch 4 "Moving Charges and Magnetism", p.150_
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