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.

Answer

`(5mu_0Itheta)/(24pir)`

Explanation

We know that magnetic field induction at the centre of an arc subtending an angle `theta` is
`B=(mu_0)/(4pi)I/rtheta`
`:.` Total magnetic field induction at O due to current through the three arcs is
`B=(mu_0)/(4pi)I[1/r-(1)/(2r)+(1)/(3r)]theta=(5mu_0Itheta)/(24pir)`