A current of `10 A` flows around a closed path in a circuit which is in the horizontal plane as shown in the figure. The circuit consists oif eight alternating arcs of radii `r_1=0.08m` and `r_2 =0.12m`. Each subtends the same angle at the centre.

a. Find the magnetic field produced by this circuit at the centre.
b. An infinitely long straight wire carryin as current of `10 A` is passing through the centre of the above circuit vertically with the direction of the current being into the pane of the circuit. what is the force acting on the wire at the centre due to the current in the circuit? What is the force acting on the arc `AC` and the straight segment `CD` due to the current at the centre?

Explanation

For finding the magnetic field produced by this circuit at the centre we can consider. It to contain of two semicircles of radius `r_(1)=0.08m` and `r_(2)=0.12m` . Since current is flowing in the same direction the magnetic field created by circular arcs will be in the same direction and hence added
`:. B_(1)=(mu_(0)i)/(4r_(1)) ` and `B_(2)=(mu_(0)i)/(4r_(2))`
`:. B=(mu_(0)i)/(4)[(1)/(r_(1))+(1)/(r_(2))](` Direction outwards )
`:. B=(6.54xx10^(-5))T` ( Right hand thumb rule )
`(b)` Force acting on a current carrying conductor plaed in a magnetic field is given by
`F=I(lxx B ) IlBsin theta`
In this case `theta 0 :. F=0`
`(ii)` On arc AC due to current at the centre `|vec(B)|` at AC will be `B=(mu_(0)I)/(2pir_(1))` . The direction of this magnetic field on any small segment of AC will be tangential `:. theta =180^(@) :. F=0`
`(iii)` On segment `CD`
Force on a small segment `dx` distant r from O
`dF=IdxB=10xxdxxx(mu_(0)I)/(2pix)`
`:. =(5mu_(0)I)/(pi)(dx)/(x)` On integrating
`:. F=(5 mu_(0)I)/(pi)int _(r_(1))^(r_(2))(dr)/(r):. F=(5mu_(0)I)/(pi)[log_(e)r]_(r_(1))^(r_(2))`
`:. F=(5mu_(0)I)/(pi)log_(e).(r_(2))/(r_(1))(5mu_(0)xx10)/(pi)log_(e)((0.12)/(0.08))=8.1xx10^(-6)N`