An electric field line emerges from a positive point charge `+q` at angle `alpha` to the straight line connecting it to a negative point charge `-2q` as shown in figure. At what angle `beta` with the field line enter the charge`-2q` ?

Options

A. `alpha`
B. `2 sin^(-1)((1)/(sqrt(2))sin.(alpha)/(2))`
C. ` sin^(-1)((1)/(sqrt(2))sin.(alpha)/(2))`
D. ` sin^(-1)((1)/(2sqrt(2))sin.(alpha)/(2))`

Answer

D

Explanation

Torque due to `vec(B)` = Torque of weight of (QRST) (about QT)
`L(B_(0)IL)=3(lambdaL)g((L)/(2)+(L)/(6))=3lambdaLg((2L)/(3))`
`B_(0)=(2lambdag)/(I)`
No external force, so COM cannot displace initial coordinate of `COM=(3(lambdaL)("zero")+2lambdaL((L)/2)+lambdaL(L))/(6lambdag)=(L)/(3)`
Final coordinate of `COM=(L)/(3)("same")`
But COM displaces with respect to QT by `(2L)/(3)`. So displacement of `QPUT=(2L)/(3)`.
Initial magnetic dipole moment M makes an angle of `(pi)/(4)` ACW from +ive x-axis and finally it makes angle `(3pi)/(4)` ACW from `+"ive x-axis" (M=sqrt(2)IL^(2))`
So change in `PE=-MB(cos theta_(2)-costheta_(1))=MB_(0)[cos135^(@)-cos45^(@)]=2B_(0)IL^(2)="Gain in KE"=(1)/(2)I omega ^(2)`
`I("about QT")=2[(2lambdaL^(3))/(3)+lambdaL^(3)]=(10lambdaL^(3))/(3),2B_(0)IL^(2)=(1)/(2)(Iomega^(2))=(1)/(2)((10lambdaL^(3))/(3))omega^(2),omega^(2)=(6B_(0)I)/(5lambdaL)`