Find the magtiude and direction of a force vector acting on a unit length of a thin wire, carrying a current `I = 8.0 A`, at a point `O`. If the wire is hent as shown in (a) Fig with curvature radius `R = 10 cm`
(b ) Fig the distance between the long parallel segments of the wire being equal to `l = 20 cm`.

Explanation

(a) The magnetic field at `O` is only due to the curved path, as for the line element, `d vec(l) uarr uarr vec(r)`.
Hence, `vec(B) = (mu_(0) i)/(4pi R) pi (-vec(k)) = (mu_(0) i)/(4R) (-vec(k))`
Thus, `vec(F_(m)) = iB (-vec(j)) = (mu_(0) i^(2))/(4R) (-vec(j))`
So, `F_(u) = (pi_(0) i^(2))/(4R) = 0.20 N//m`
(b) In this part, magnetic induction `vec(B)` at `O` will be effective only due to the two semiinfnite segments of wire. Hence
`vec(B) = 2. (mu_(0)i)/(4pi ((l)/(2))) "sin" (pi)/(2) (-vec(k))`
`= (mu_(0) i)/(pi i) (-vec(k))`
Thus force per unit length,
`vec(F_(u)) = (mu_(0) l^(2))/(pi l) (-vec(i))`