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A very long cylindrical wire is carrying a current $I_{0}$ distributed uniformly over its cross-section area. O is the centre of the cross-section of the wire and the direction of current into the plane of the figure. The value of $\int_{A}^{B} \vec{B} . \vec{d I}$ along the path AB (from A to B) is

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$μ_{0} I_{0}$

$\frac{μ_{0} I_{0}}{3}$

$- \frac{μ_{0} I_{0}}{6}$

$\frac{μ_{0} I_{0}}{6}$

answer is B.

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Detailed Solution

According to Ampere’s circuital law
$\oint \vec{B} . \vec{d I} = μ_{0} I_{n e t}$
For the closed path OAB
$\oint \vec{B} . \vec{d l} = \int_{O}^{A} \vec{B} . \vec{d l} + \int_{B}^{O} \vec{B} . \vec{d l} = μ_{0} (- \frac{I_{0}}{6})$

Since the magnetic field is perpendicular to the paths OA and BO

$\int_{A}^{B} \vec{B} . \vec{d l} = \frac{- μ_{0} I_{0}}{6}$

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