Three long straight wires P, Q and R cross the plane of the paper at right angles. P carries 5 A out of the paper and Q carries 2 A into the paper. Both lie inside a closed loop drawn on the paper. R carries 3 A out of the paper and lies outside the loop. Going anticlockwise round the loop, as seen by a person looking at the paper, the value of \(\oint\vec{B}\cdot d\vec{l}\) (in SI units) is
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \(3\mu_0\)
Ampere’s circuital law gives \( \oint\vec{B}\cdot d\vec{l}=\mu_0I_e \). Here \( I_e \) is the net current that passes through the loop. A current outside the loop does not count, even though its field acts all along the loop.
The sign of each current comes from the right-hand rule. Curl the fingers of the right hand in the sense of going round the loop. For an anticlockwise path on the paper, the thumb points out of the paper. So a current out of the paper is positive, and a current into the paper is negative.
\( \begin{aligned} I_e &= 5\ \mathrm{A}-2\ \mathrm{A} \\ &= 3\ \mathrm{A} \end{aligned} \)
\( \begin{aligned} \oint\vec{B}\cdot d\vec{l} &= \mu_0I_e \\ &= \mu_0\times3 \\ &= 3\mu_0 \end{aligned} \)
Wire R, carrying 3 A, lies outside the loop, so it adds nothing to the loop integral.
NCERT (2026-27):
Ampere’s Circuital Law – “… Ampere’s law states that this integral is equal to μ0 times the total current passing through the surface, i.e.,”
\( \displaystyle \oint\vec{B}\cdot d\vec{l}=\mu_0 I \)\( \qquad [4.13(a)] \)
Ampere’s Circuital Law – “… Let the fingers of the right-hand be curled in the sense the boundary is traversed in the loop integral ∮B.dl. Then the direction of the thumb gives the sense in which the … current I is regarded as positive.”