Two identical current carrying coaxial loops, carry current I in an opposite sense. A simple amperian loop passes through both of them once. Calling the loop as C ______. a. oint B.dl = +- 2μ_0I b. the value of oint B.dl is independent of sense of C. c. there may be a point on C where B and dl are perpendicular. d. B vanishes everywhere on C.

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b and c Explanation: Ampere’s law gives another method to calculate the magnetic field due to a given current distribution. Line integral of the magnetic field vecB around any closed curve is equal to μ0 times the net current i threading through the area enclosed by the curve, i.e. `oint vecB * vec(dl) = mu_0 sumi = mu_0 (i_1 + i_3 - i_2)` Total current crossing the above area is (i_1 + i_3 - i_2). Any current outside the area is not included in net current. (Outward ⊙ $\to$ + ve, Inward ⊗ $\to$ – ve) Applying the Ampere's circuital law, we have ointB* dl = i_0 (I - I) = 0 (because current is in opposite sense) Also, there may be a point on C where B and dl are perpendicular and hence, oint_c B*dl = 0 [Solution diagram: 326725_solution_1.png]

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