In the given figure net magnetic field at O will be

Options

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

Answer

B

Explanation

Inner loop ,
`B_(1) = (mu_(0))/(4pi) ((pi i))/((a//2)) otimes " "`(-ve z-axis)
Outer loop
`B_(2) = (mu_(0))/(4pi) ((pi i)/(3a//2)) odot` (+ ve z-axis)
Along z-axis
`B_(1) - B_(2) = (mu_(0))/(4pi ) * (pi i)/(a) [2 - (2)/(3)] otimes`
Wire-inner ,
`B_(3) = (mu_(0))/(4pi ) * (i)/((a//2))` (-ve y-axis)
Outer ,
`B_(4) = (mu_(0))/(4 pi ) ((i)/(3a//2))` (-ve y-axis)
`B_(3) + B_(4) = (mu_(0))/(4pi) . (I)/(a) (2 + (2)/(3))` (-ve y-axis)
`B_("net") = sqrt((B_(1) + B_(2))^(2) + (B_(3) + B_(4))^(2)) = (mu_(0) i)/(3pi a) sqrt(pi^(2) + 4)`