emf = BLV

= 1 . (2R) . 1

= 2V

$$r = {P \over {qB}} = {{\sqrt {2mk} } \over {qB}}$$

Given they have same kinetic energy

$$r \propto {{\sqrt m } \over q}$$

$${{{r_1}} \over {{r_2}}} = {{\sqrt 4 } \over 2} \times {3 \over {\sqrt {16} }} = {3 \over 4}$$

$${r_2} = {{4{r_1}} \over 3}$$ (r₂ is for heavier ion and and r₁ is for lighter ion)


$$\sin \theta = {d \over R}$$

$$\theta$$ $$\to$$ Deflection

$$\theta \propto {1 \over R}$$

(R $$\to$$ Radius of path)

$$\because$$ R₂ > R₁ $$\Rightarrow$$ $$\theta$$₂ < $$\theta$$₁

$$\to$$ When magnet passes through centre region of solenoid, no current / Emf is induced in loop.

$$\to$$ While entering flux increases so negative induced emf

$$\to$$ While leaving flux decreases so positive induced emf.

$${{{E_0}} \over C} = {B_0}$$

$${F_{\max }} = e{B_0}V$$

$$ = 1.6 \times {10^{ - 19}} \times {{800} \over {3 \times {{10}^8}}} \times 3 \times {10^7}$$

$$ = 12.8 \times {10^{ - 18}}$$ N