Question

The region between y = 0 and y = d contains a magnetic field vector B = Bz . A particle of mass m and charge q enters the region with a velocity v = v i . If d = mv/2qB, the acceleration of the charged particle at the point of its emergence at the other side is______

Answer 1

Given:

The region between y = 0 and y = d contains a magnetic field vector B = B(z) . A particle of mass m and charge q enters the region with a velocity v = v i , d = mv/2qB.

To find:

Acceleration at the time of leaving the field.

Calculation:

Usual radius of the circular Trajectory for charge

$\therefore \: r = \dfrac{mv}{qB}$

But, the dimensions of the field is half that of the radius of circular trajectory.

$\therefore \: r = \dfrac{mv}{qB} = 2d$

Now , let deviation be $\theta$.

$\sin( \theta) = \dfrac{d}{2d} = \dfrac{1}{2}$

$= > \theta = {30}^{ \circ}$

So, net acceleration vector :

$\vec{a} = \dfrac{qvB}{m} \bigg \{ \cos( \theta) ( - \hat{i}) + \sin( \theta) \hat{j} \bigg \}$

$= > \vec{a} = \dfrac{qvB}{m} \bigg \{ \cos( {30}^{ \circ} ) ( - \hat{i}) + \sin( {30}^{ \circ} ) \hat{j} \bigg \}$

$= > \vec{a} = \dfrac{qvB}{m} \bigg \{ \dfrac{ \sqrt{3} }{2} ( - \hat{i}) + \dfrac{1}{2} \hat{j} \bigg \}$

$= > \vec{a} = \dfrac{qvB}{m} \bigg \{ - \dfrac{ \sqrt{3} }{2} \hat{i} + \dfrac{1}{2} \hat{j} \bigg \}$

So, final answer is:

$\boxed{ \sf{ \red{\vec{a} = \dfrac{qvB}{m} \bigg \{ - \dfrac{ \sqrt{3} }{2} \hat{i} + \dfrac{1}{2} \hat{j} \bigg \}}}}$