Magnetics
6.0 Deviation of charged particle in magnetic field
6.0 Deviation of charged particle in magnetic field
Suppose a charged particle $q$ enters normally in a uniform magnetic field $\vec B.$ The magnetic field extends to a distance $x,$ which is less than or equal to radius of the path, that is $x \leqslant r.$
i. For $x \leqslant r$
The radius of path $$r = \frac{{mv}}{{qB}}$$And $$\sin \theta = \frac{x}{r}$$ii. For $x>r$$$r = \frac{{mv}}{{qB}}$$and deviation, $θ = 180^\circ$ as clear from the diagram.
iii. If particle moves for time t inside the field, then$$\theta = \omega t$$$$\theta = \left( {\frac{{Bq}}{m}} \right)t$$
As, $$\alpha = \frac{q}{m}$$$$\therefore \quad \theta = B\alpha t$$
iv. Velocity of particle $:-$
We have $$\theta = {B_0}\alpha t,$$ $$r = \frac{{{v_0}}}{{\alpha {B_0}}}$$Velocity of particle at any time $t,$$$\vec v = {v_x}\hat i + {v_y}\hat j$$$${\vec v_0} = {v_0}\cos \theta \;\hat i + {v_0}\sin \theta \;\hat j$$On substituting the value of $\theta,$ we have
Or$$\vec v = {v_0}\cos \left( {{B_0}\alpha t} \right)\;\hat i + {v_0}\sin \left( {{B_0}\alpha t} \right)\;\hat j$$
v Position of particle $:-$$$\vec r = x\hat i + y\hat j$$$$\vec r = r\sin \theta \;\hat i + \left( {r - r\cos \theta } \right)\hat j$$$$\vec r = r\left[ {\sin \theta \;\hat i + 1\left( {r\cos \theta } \right)\hat j} \right]$$$$\vec r = \frac{{{v_0}}}{{{B_0}\alpha }}\left[ {\sin \left( {{B_0}\alpha t} \right)\hat i + \left\langle {1 - r\cos \left( {{B_0}\alpha t} \right)} \right\rangle \hat j} \right]$$
