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In the given figure net magnetic field at O will be |
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Answer» `(2mu_(0)i)/(3pi a) sqrt(4-PI^(2))`  Magnetic FIELD at 0 due to Part(1) : `B_(1)= 0` Part(2): `B_(2) = (mu_(0))/(4pi).(pi i)/((a//2)) ox` (along-Z-axis) Part(3) `B_(3) = (mu_(0))/(4pi) .(i)/((a//2))(darr)` (along-Y-axis) Part (4): `B_(4) = (mu_(0))/(4pi).(pi i)/((3a//2)) o.` (along +Z-axis) Part (5): `B_(5) =(mu_(0))/(4pi).(i )/((3a//2))(darr)` (along-Y-axis) (along-Z-axis) `B_(3) + B_(5) = (mu_(0))/(4pi).(1)/(a) (2 + (2)/(3))= (8mu_(0)i)/(12pi a) (darr)` (along-Y-axis) Hence net magnetic field `B_("net") = sqrt((B_(2)- B_(4))^(2) +(B_(3)+ B_(5))^(2)) = (mu_(0)i)/(3pi a) sqrt(pi^(2) + 4)` |
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