Two long straight wires enter a room through a window. one carries a current of 2.9a into the room, while the other carries a current of 4.4 out of the room. calculate the magnitude (in t. m) of the path integral of b. ds around the window frame.

Answer and Explanation:

curents i = 2.9 A

           i ' = 4.4 A

the magnitude (in T.m) of the path integral of B.dl around the window frame = μo * current enclosed

          = μo* ( i '- i )

Since from Ampere's law

where μ o = permeability of free space = 4π * 10 ^-7 H / m

plug the values we get the magnitude (in T.m) of the path integral of B.dl = ( 4π*10^-7 ) (2.9+4.4)

                                 = 1.884 * 10^-6 Tm

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