Foolish $\alpha$ Particle in Variable Magnetic Field

The speed of particle remains constant since it is a magnetic field and magnetic field exerts force always along a direction perpendicular to velocity , hence at any instant , The radius of curvature is given by

$\frac { mv }{ qBx }$

now,, when it passes through a thin element dx

it deviates by

a small angle d(theta)

given by

$d(sin(\theta ))=\frac { dx }{ R }$

substituting the value of R we get

$d(sin(\theta ))=\frac { qBxdx }{ mv }$ .......(1)

Now when it reaches the maximum distance along x-axis, Its velocity has no component along x-axis or it is perpendicular to x-axis ,,

hence ,, from the beggining ,, it has deviated by an angle of 90 degrees

so integrating eqn ... (1) from 0 to pi/2 for RHS and from 0 to 'a' for LHS

We get

$1=\quad \frac { qB{ a }^{ 2 } }{ 2mv } \\ \\ or\quad a\quad =\quad \sqrt { \frac { 2mv }{ qB } }$

Now using the formula for R

we get

$\\ R=\quad \sqrt { \frac { mv }{ 2qB } }$

which is nothing but

$2\sqrt { 2 } =\quad 2.82847125$

and why is the alpha particle foolish ?

i did it using force approach.

Let particle has velocity v and u at any instant. (v along x and u along y)

Writing force equations

-qBxv/m = du/dt = udu/dy

qBxu/m = dv/dt = vdv/dx

Also using energy conservation v^2 + u^2 = (vo)^2

Substitute value of u in first equation and integrate from 0 to x and velocity from (vo) to zero as at maximum x coordinate velocity will be zero. (x velocity obviously)

Now just apply formula of radius of curvature to get required answer