Solution:
current enclosed is (1+2+3-1-4)= 1A
using ampere circuital law we get the answer.
Six wires of current I1 = 1A, I2 = 2A, I3 = 3A, I4 = 1A, I5 = 4A, I6 = 5A cut the page perpendicular at points 1,2,3,4,5 and 6. The value of time integral of \(\overrightarrow{B}\) around the dotted closed path \(\left( i.e, \oint_{}^{}\overrightarrow{B} \overrightarrow{dl} \right)\) is :
Leave a Reply