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Exercise · Q16

Q.State Ampere's circuital law in integral form and compare it, as a calculation tool, with Gauss's law of electrostatics.

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Ampere's circuital law states that the line integral of B⃗\vec{B} around any closed Amperian loop equals μ0\mu_0 times the total current enclosed:

∮B⃗⋅dl⃗=μ0 Ienc\oint \vec{B}\cdot d\vec{l} = \mu_0\, I_{\text{enc}}

with sign fixed by curling the right hand's fingers along the loop's traversal direction, thumb giving the positive current sense. It plays exactly the role Gauss's law plays for E⃗\vec{E}: both are exact relations between a field and its source, and both become genuinely useful calculation shortcuts only when a symmetric loop (or surface) can be chosen so the field is constant in magnitude and eit …

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