A thin wire `AC` shaped as a semi-circle of diaameter `d = 20 cm` rotates with a constant angular velocity `omega = 100` rad/s in a unifrom magnetic field of induction `B = 5.0 mT`, with `omega uarr uarr B`. The rotation axis passes through the end `A` of the wire and is perpendicular to the diameter `AC`. Find the value of a line intergal `int E dr` along the wire from point `A` to `C`. Generalize the botained result.

Explanation

By definition,
`vec(E) = -(vec(v) xx vec(B))`
So, `int_(A)^(C) vec(E).d.vec(r) = int_(A)^(C) -(vec(v) xx vec(B)).d vec(r) = int_(0)^(d) -v B dr`
But, `v = omega r`, where `r` is the perpendicular distance of the point from `A`.
Hence, `int_(A)^(C) vec(E).d.vec(r) = int_(0)^(d) -omegab B r dr = -(1)/(2) omega B d^(2) = -10 mV`
This result can be gencralized to a wire `AC` of orbitary planner shape. We have
`int_(A)^(C) vec(E).d.vec(r) = -int_(A)^(C) (vec(v) xx vec(B)). dr = -int_(A)^(C) ((omega xx vec(r)) xx vec(B)). d vec(r)`
`= -int_(A)^(C) (vec(B).vec(r).vec(omega) - vec(B).vec(omega).vec(r)).d vec(r)`
`= - (1)/(2) B omega d^(2)`,
d being `Ac` and `vec(r)` being measured from `A`.