A uniform wire of resistance per unit length `lambda` is bent into semicircle of radius `a`. The wire rotates with angular velocity `omega` in a vertical plane about a orizontal is passing through `C`. A uniform magnetic field `B` exists in space in a direction perpendicular to paper inwards.

a. Calculate potential difference between points `A` and `D`. which point is at higher potential?
b. If points `A` and `D` are connected by a conducting wire of zero resistance, find the potential difference between `A` and `C`.

Explanation

a. Length of straight wire `AC` is `l_1=2asin(theta/2)`

Therefore, the motional emf (or potential difference) between points `C` and `A` is
`V_(CA)=V_C-V_A=1/2Bomegal_1^2=2a^2Bomegasin^2(theta/2)` ..(i)
From right hand rule, we can see that `V_Cgt V_A`
Similarly length of straight wire `CD` is
`l_2=2asin(pi/2-theta/2)=2cos(theta/2)`
Therefore the `PD` between points `C` and `D` is
`V_(CD)=V_C-V_D=1/2 B omegal_(2)^(2) = 2alpha^(2) B omega cos^(2) ((theta)/(2))` ...(ii)
with `V_(C) gt V_(D)`
Eq. (ii) -Eq.(i) gives
`V_(A) -V_(D) = 2a^(2)B omega (cos^(2)(theta)/(2)-sin^(2)(theta)/(2))`
`= 2a^(2) B omega cos theta`

`A` is at higher potential.
(b) When `A` and `D` are connected from a wire current starts flowing in the circuit as shown in figure:
Resistance between `A` and `C` is `r_1=`(length of are `AC` ) `lambda=atheta lambda`
and between `C` and `D` is `r_2=`(length of are `CD`) lambda=(pi-theta)alambda`

In the figure `E_1=2^2Bomegasin(theta/2)` and `E_2=2a^2Bomgacos^2(theta/2)`
with `E_2gtE_1`
`:.` Current in the circuit is
`i=(E_2-E_1)/(r_1+r_2)=(2a^2Bomegacostheta)/(pi a lambda)=(2aBomegacostheta)/(pi lambda)`
and potential difference between points `C` and `A` is
`V'_(CA) =E_1+ir_1=2a^2Bomegasin^2(theta/2)+((2aBomegacostheta)/(pi lambda))(athetalambda)`
`=2a^2Bomega(sin^2theta/2+theta/picostheta)`