Question

In the given figure net magnetic at `O` will be

A. `(2mu_(0)i)/(3pi a) sqrt(4 - pi^(2))`
B. `(mu_(0)i)/(3pi a) sqrt(4 + pi^(2))`
C. `(2mu_(0)i)/(3pi a^(2)) sqrt(4 +pi^(2))`
D. `(mu_(0)i)/(3pi a) sqrt((4 + pi^(2)))`

Answer 1

Correct Answer - B

Magnetic field at `O` 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). (pi i)/((a//2)) (darr)` (along `Y` axis)
Part(4): B_(4) `= (mu_(0))/(4pi),. (pi i)/((3a//2)) o.` (along `+Z` axis)
Part(5)L `B_(5) =B (mu_(0))/(4pi) .(i)/((3a//2)) (darr)` (along `-Y` axis)
`B_(2) - B_(4) = (mu_(0))/(4pi), (pi)/(a) (2 - (2)/(3)) = (mu_(0) i)/(3a) ox` (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)`