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(D) According to Ampere's circuital law,the line integral of the magnetic field $\vec{B}$ around any closed loop is equal to $\mu_0$ times the net cur

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(D) According to Ampere's circuital law,the line integral of the magnetic field $\vec{B}$ around any closed loop is equal to $\mu_0$ times the net current $I_{	ext{enclosed}}$ passing through the surface bounded by the loop.$\oint \vec{B} \cdot d\vec{l} = \mu_0 I_{	ext{enclosed}}$In this problem,the circular path of radius $R$ does not enclose the straight wire carrying current $I$. Therefore,the net current enclosed by the circular path is $I_{	ext{enclosed}} = 0$.Since the integral is taken over the closed circular path,and the magnetic field produced by the external wire is conservative in the region not containing the wire,the line integral of the tangential component $\vec{B}_T$ around the closed loop is zero.Thus,$\oint \vec{B}_T \cdot d\vec{l} = 0$.

In the following diagram,there is a straight wire carrying a current $I$. Consider a circular path with radius $R$ near it. If $\vec{B}_T$ is the tangential component of the magnetic field along the circular path,then find the value of the integral $\oint \vec{B}_T \cdot d\vec{l}$.

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