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) The `beta` - particle is passing through both electric and magnetic fields and suffers no deviation i.e., it goes straight through region I where both `vec E " and " vec B` are present. It means that the force on `beta` particle due to electric field `( - e vec E )`, acting upwards is equal and opposite to the force due to magnetic field which is `- e (vec v xx vec B)` , acting downwards.
Using Fleming.s left hand rule, the direction of magnetic field `(vec B)` is perpendicular to the plane of paper and inward.
(ii) Force due to electric field = Force due to magnetic field
`eE = evB sin 90^(@)`
` eE = evB`
`v = E/B`
(iii) Let r be the radius of circular path of `beta` particle
in region II.
Then ` evB = (mv^(2))/r`
` r = (mv)/(eB)`
If magnetic field is maden times i.e.
` B. = nB`
The new radius , `r. = (mv)/(enB) = 1/n (mv)/(eB)`
`rArr " " r. = 1/n r`