A fine pencil of `beta`-particles, moving with a speed v, enters a region (region I), where a uniform electric field and a uniform magnetic field are both present. These `beta`-particles then move into region II where only the magnetic field, (out of the two fields present in region I) exists. The path of the `beta`-particles, in the two regions is as shown in the figure.

(i) state the direction of magnetic field.
(ii) state the relation between E and B in region I.
(iii) Drive the expression for the radius of the circular path of the `beta`-particle in region II.
(iv) If the magnitude of magnetic field, in region II is changed to n times its earlier value, (without changing the magnetic field in region I) find the factor by which the radius of this circular path would change.

Explanation

(i) As the `beta`-particle goes undeviated while passing through the electric and magnetic fields, so the force on moving `beta`-particle due to electric field `(-evecE)` must be equal and opposite to the force due to magnetic field `[-e(vecvxxvecB)]`. Due to it, the direction of `vecB` must be perpendicular to plane of paper downwards.
(ii) In region I as `beta`-particle goes undeviated
`-evecE=-e(vecvxxvecB)` or `eE=evB` or `v=E//B`
(iii) If r is the radius of circular path of the `beta`-particle in region II, then
`evB=mv^2//r` or `r=(mv)/(eB)`
(iv) If `B^'=nB`, then `r^'=(mv)/(eB^')=(mv)/("ne"B)`
`:. r^'/r=1/n` or `r^'=r/n`