six wires of current I1 = 1A , I2 = 2A, I3 = 3A , I4 = 1A , I5 = 5A and I6 = 4A cut the page perpendicularly the points 1,2,3,4,5 and 6 respectively as shown in the figure. Find the value of the integral B?dl around the closed path.

Dear student,

From Ampere circuital law , the line integral of magnetic field around the closed loop is equal to μ0 times the algebraic sum of current passing through loop.

B.dl =μ0.I

As wire 5 is outside loop hence we have to consider other 5 wires.
Net current of other wire is 
Hence the value is given by
B.dl =μo×1B.dl =μo


  • 5
Hello Yaman dear, we have to apply Ampere's circuital law ​∫ B;dl = µo * net current passing through the closed loop
Wire 5 carrying current 5 A is not to be taken into account as it is outside of the loop
For the other five wires currents are in opposite directions
So algebraic sum of cuuents is needed
Net current = (1 + 2 + 3 - 1 - 4) = µo * 1 weber m^-1
Hence the required closed ∫ B;dl = ​µo weber m^-1

  • 2
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