when x < a

$${B_1}(2\pi x) = {\mu _0}\left( {{{{i_0}} \over {\pi {a^2}}}} \right)\pi {x^2}$$

$$B(2\pi x) = {{{\mu _0}{i_0}{x^2}} \over {{a^2}}}$$

$${B_1} = {{{\mu _0}{i_0}x} \over {2\pi {a^2}}}$$ .... (1)

when a < x < b

$${B_2}(2\pi x) = {\mu _0}{i_0}$$

$${B_2} = {{{\mu _0}{i_0}} \over {2\pi x}}$$ ..... (2)

$${{{B_1}} \over {{B_2}}} = {{{\mu _0}{i_0}{x \over {2\pi {a^2}}}} \over {{{{\mu _0}{i_0}} \over {2\pi x}}}} = {{{x^2}} \over {{a^2}}}$$

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.