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A very long straight conducting wire, lying along the z-axis, carries a current of 2 A. The integral $\oint \vec{B} \cdot d \vec{l}$ computed along the straight line PQ, where P has the coordinates (2 cm, 0, 0) and Q has the coordinates (2 cm, 2 cm, 0). The integral has the magnitude (in SI units)

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$\frac{π}{2} \times 10^{- 7}$

$8 π \times 10^{- 7}$

$2 π \times 10^{- 7}$

$π \times 10^{- 7}$

answer is D.

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Detailed Solution

$\int_{P A Q} \vec{B} \cdot d \vec{l} = \int_{R B Q} \vec{B} \cdot d \vec{l} (∵ \int_{P C R} \vec{B} \cdot d \vec{l} = 0) = (\frac{μ_{0} I}{2 π R}) (\frac{π}{4} R) = \frac{(4 π \times 10^{- 7}) (2)}{8} = π \times 10^{- 7} S.I. units$

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