Question

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 1

(a) Magnetic field at the center due to the straight parts is
zero.
Magnetic fild due to the arcs of radius `r_1:`
`=4(mu_0i//2r_1)(1//8)=(mu_0i//4r_1)`
Similarly, the magnetic field due to the arcs of radius `r_2`:
`=(mu_0i//4r_2)`
`implies "Net magnetic field"`
`=(mu_0i//4){(1//r_1+1//r_2}=6.54xx10^-5T`
(b) (i) As the current in the wire at the centre is antiparallel
to the direction of magnetic field, the force on the wire
everywhere will be zero.
(ii) Further due to the current at the centre, the magnetic
field at AC will be tangential and hence the force on
AC will be zero.
(iii). Force on `CD=int_(r_1)^(r_2) (mu_0i)/(2pix) idx=(mu_0i^2)/(2pi) In ((r_2)/(r_1))`
`=8.11xx10^-6N ("Vertically downward")`