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

(a) `B_("centre") = (mu_(0)i)/(2r_(1)) xx (pi//4)/(2pi) xx 4 + (mu_(0)i)/(2r_(2)) xx (pi//4)/(2pi) xxx 4`
`= (mu_(0)i)/(4)((1)/(r_(1)) + (1)/(r_(2)))`
`=(4pi xx 10^(-7) xx 10)/(4) ((1)/(0.08) + (1)/(0.12))`
`= pi xx 10^(-6) ((12 + 8)/(0.96))`
`= pi xx 10^(-6) xx (20)/(0.96) = (pi)/(0.96) = (pi)/(48) xx 10^(-3)`
`= 6.54 xx 10^(-5)T, o.`
(b) Since current in wire is anti-parallel to magnetic field, hence force on wire is zero.
Arc `AC`:
Here magnetic field due to long wire at centre `C` is antiparallel to current.
`F_(AC) = 0`
Arc `CD` :
Magnetic field due to long wire at distance `x`
`B = (mu_(0)i)/(2pi x), _|_^(ar)` to wire `CD`, as shown
Force on small length `dx`
`dF = Bidx`
`F_(CD) = (mu_(0)i^(2))/(2pi)int_(r_(1))^(r_(2)) (dx)/(x)`
`= (mu_(0)i^(2))/(2pi)|log_(e) x|_(r_(1))^(r_(2))`
`= (mu_(0)i^(2))/(2pi)|log_(e)r_(2) - log_(e)r_(1)|`
`= (mu_(0)i^(2))/(2pi) log_(e)((r_(2))/(r_(1)))`
`= 2 xx 10^(-7) xx (10)^(2) log_(e) ((0.12)/(0.08))`
`= 2 xx 10^(-5) log_(e) (1.5)N`, vertically downward