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Ampere's Circuital Law states that around any CLOSED loop, the line integral of $\vec{B}\cdot d\vec{l}$ equals
Athe enclosed current divided by $\varepsilon_0$
B$\mu_0$ times the enclosed current
Cthe total magnetic flux through the loop
Dthe average magnetic field times the loop perimeter
Answer & Solution
Correct answer: B. $\mu_0$ times the enclosed current
1. NCERT §4.5 states Ampere's Law: $\oint\vec{B}\cdot d\vec{l} = \mu_0\,I_\text{enc}$.
2. The LEFT side is the line integral of $\vec{B}$ around any CLOSED loop (Amperian loop).
3. The RIGHT side is $\mu_0$ times the TOTAL CURRENT passing through any surface bounded by the loop.
4. Sign convention: use the right-hand rule on the loop's orientation — current parallel to thumb is positive.
5. Same as Gauss's law in spirit: useful when SYMMETRY (cylindrical, planar, toroidal) lets you pull $B$ outside the integral.
6. Option A is Gauss's law for electricity. Options C and D have wrong functional forms.
_Source: NCERT Class 12 Physics Part 1, Ch 4, §4.5 (Ampere's Circuital Law, Eq. 4.20), p. 14–15._
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