A current of `10 A` flows around a closed path in a circuit which is in horizontal plane as shown in the figure . The circuit consists of eight alternating arcs of ` radii r_(1) = 0.08 m and r_(1) = 0.12 m`. Each arc subtends the same at he center.
(a) Find the magnetixc field produced by this circuit at the center.
(b) An infinitely long straight wire carrying a current of `10 A` is passing through the center of the above circuit vertically with the direction of the current being into the plane of the circuit . What is the force acting on the arc `AC` and the straight segment ` CD` due to the current at the center?

Answer

A, B, D

Explanation

For finding the magnetic field produced by this circuit at the centre we can consider it to contain two semicircles of radii, ` r_(1) = 0.08 m and r_(2) = 0.12 m` . Since currents is flowing in the same direction , the magnetic field created by circular arcs will be in the same direction and therefore will be added.
:. `vec(B)_(1) =(mu_(0) i)/( 4 r_(1) and B_(2) = (mu_(0)i)/( 4 r_(2)) :. B = ( mu_(0)i)/(4) [(1)/(r_(1)) + (1)/( r_(2))]`
:. `B = ( 6.54xx10^(-5))T` Directed outwards.
( right hand thumb rule)
(b) Force acting on a current carrying conductor placed in a magnetic field i is given by `vec(F) = I((vec(l)xx vec(B)) = I lB sin theta `
(i) For force acting on the wire at the centre
in this case `theta = 180^(@)`
:. `F = 0`
(ii) On arc ` AC` due to current at the centre
`|vec(B)|` on `AC` will be `B = (mu_(0)I)/( 2 pi r_(1))`
The direction of this magnetic field on any small segment of `AC` will be tangential
:. `theta = 180^(@) rArr F = 0`
(iii) On segment `CD`.
Force on a small segment ` dx` distant `r` from `O`
` dF = I dxB`
` = 10 xx dx xx(mu_(0)I)/(2 pi x) = (5 mu_(0)I)/(pi) ( dx)/(x)`
On integrating
`F = ( 5 mu_(0)I)/(pi) int_(r1)^(r2) (dx)/(x) :. F = ( 5 mu_(0)I)/(pi)[ log_(e)x]_(r1)^(r2)`
`F = ( 5 mu_(0)I)/(pi)(log_(e)) (r_(2))/(r_(1)) = ( 5 mu_(0)xx10)/(pi) log_(e) ((0.12)/(0.08))`
` = 8.1 xx10^(-6)N`
directed downwards ( By Fleming left hand rule).