A conductor carrying current I is of the type as shown in figure. Find the magnetic field induction at the common centre O of all the three arcs.

Options

A. `(3 mu_(o)I theta)/(4 pi r)`
B. `(5 mu_(0)I theta)/(24 pi r)`
C. `(mu_(o)I theta)/(8 pi r)`
D. `(mu_(o) I theta)/(pi r)`

Answer

B

Explanation

We know that
`B = (mu_(o)I)/(4 pi r)theta` due to the are
`therefore` Total magnetic field induction at O, due to current through the three are is
`B = (mu_(o)I)/(4 pi) theta [(1)/(r) - (1)/(2r) + (1)/(3r)]`
`B = (5 mu_(o)I theta)/(24 pi r)`