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?

Answer

(a) `6.6 xx 10^(-5) T, (b) 0 , 0, 8 xx 10^(-6) Nt` .

Explanation

`B =4 xx (mu_(0))/(4pi) (I)/r_(1) xx (pi)/(4) + 4 xx (mu_(0))/(4pi) (I)/(r_(2)) xx (pi)/(4)`
`B = (mu_(0)I)/(4r_(1)) + (mu_(0)I)/(4r_(2)) = (mu_(0)I)/(4) (((1)/(r_(1)) + (1)/(r_(2)))o.`
` B =(12.57 xx 10^(-7))/(4) xx 10 xx [(1)/(0.08) + (1)/(0.12)]`
`B =3.16 xx 10^(-6) xx [(25)/(2) + (25)/(3)]`
`oversetrarrB =6.6 xx 10^(-5) jo.`
(b) Force on the wire at center wire be zero because current in wire will be anti parallel to field at centre
(ii) Force on arc`AC` will also zero becuase field due to wire at `AC` will be tangent to current at each point
`F_(CD) = underset(0.08)overset(0.12)intIB_(x) dx = underset(0.08)overset(0.12)intI(mu_(0))/(4pi) (2I)/(x) dx = (2mu_(0)I^(2))/(4pi)`
` underset(0.08)overset(0.12)int(x)/(dx) = (2mu_(0))/(4pi) I^(2) [log_(C) x]_(0.08)^(0.12) = 2 (mu_(0))/(4pi) I^(2)`
`[log _(e) "(0.12)/(0.08)]=2 xx 10^(-7)xx 100 log_(c) 1.5 =8`
`xx 10^(-6) N`
.