Repeat above Question 3, if the charge is negative and the angle made by the boundary with the velocity is `theta = pi/6.`

Explanation

(i) `2pi-2theta=2pi-2.(pi)/(6)=2pi-(pi)/(3)=(5pi)/(3)`
`=omegat=(qBt)/(m)rArrt=(5pim)/(3qB)`

(ii) Distance travelled `s=r(2pi-2theta)=(5pir)/(3)`
(iii) Impulse = charge in linear momentum
`=m(-vsinthetahati+vcosthetahatj)-m(vsinthetahati+vcoshatj)`
`=-2mvsinthetahati=-2mvsin(pi)/(6)hati=-mvhati`