In Ampere's $\oint \vec{B} \cdot d \vec{l} = μ_{0} i,$ the current outside the curve is not included on the right hand side. Does it mean that the magnetic field B calculated by using Ampere's law, gives the contribution of only the currents crossing the area bounded by the curve?

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Solution

In Ampere's law $\oint \vec{B} . \vec{d l} = μ_{o} i$ , i is the total current crossing the area bounded by the closed curve. The magnetic field B on the left-hand side is the resultant field due to all existing currents.

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In Ampere circuital law, the current outside a curve is not included. Does it mean that magnetic field B calculated using this law gives the contribution of only currents crossing the area bounded by the curve? Explain.

Q. If

_{1},_{2}

and

_{3}

are the magnetic field due to the wires carrying current

I_{1}, I_{2}

and

I_{3}

, then in the expression of Ampere's circuital law,

\oint \to B . \to d l = μ_{0} I

\to B

Q. State Ampere's circuital law. Use it to find magnetic field due to an infinity long thin current carrying wire.

Q. Write Ampere's circuital law. A long straight wire of a circular cross-section (radius

a

) carrying steady current. Current is uniformly distributed in the wire. Calculate magnetic field inside the region

(r < a)

in the wire.

Q. Magnetic scalar potential is defined as

U ({\vec{r}}_{2}) - U ({\vec{r}}_{1}) = - \int_{{\vec{r}}_{1}}^{{\vec{r}}_{2}} \vec{B} . \vec{d l}

Apply this equation to a closed curve enclosing a long straight wire. The RHS of the above equation is then −μ₀ i by Ampere's law. We see that

U ({\vec{r}}_{2}) \neq U ({\vec{r}}_{1})

even when

{\vec{r}}_{2} = {\vec{r}}_{1}

. Can we have a magnetic scalar potential in this case?

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In Ampere's ∮B→·dl→=μ0i, the current outside the curve is not included on the right hand side. Does it mean that the magnetic field B calculated by using Ampere's law, gives the contribution of only the currents crossing the area bounded by the curve?

In Ampere's law ∮B→.dl → = μoi, i is the total current crossing the area bounded by the closed curve. The magnetic field B on the left-hand side is the resultant field due to all existing currents.

In Ampere's $\oint \vec{B} \cdot d \vec{l} = μ_{0} i,$ the current outside the curve is not included on the right hand side. Does it mean that the magnetic field B calculated by using Ampere's law, gives the contribution of only the currents crossing the area bounded by the curve?

In Ampere's law $\oint \vec{B} . \vec{d l} = μ_{o} i$ , i is the total current crossing the area bounded by the closed curve. The magnetic field B on the left-hand side is the resultant field due to all existing currents.