Magnetics
5.0 Path or Motion of charged particle in uniform magnetic field
5.0 Path or Motion of charged particle in uniform magnetic field
The path of charged particle in magnetic field depends on the angle $θ$ between $\vec v$ and $\vec B.$
Depending on different values of $θ,$ the possible cases are $:-$
Case $1 :-$ When $\theta$ is $0^\circ$ or $180^\circ\ \sin \theta = 0$
For $\theta =0^\circ$ or $180^\circ ,$ the force on the moving charge $\vec F_m=qvb(0)=0$ and therefore particle goes undeviated along a straigth path.
Case $2 :-$ When $\theta =90^\circ :-$
(i) When a particle is projected from inside the field : it experiences a force which always perpendicular to the velocity and so its path will be circular. The necessary centripetal force is provided by the magnetic force. If $r$ be the radius of the path, then$$\frac{{m{v^2}}}{r} = qvB\sin 90^\circ $$ $$r = \frac{{mv}}{{qB}}$$We can write as, $$r = \frac{{mv \bot }}{{qB}}$$$$r = \frac{v}{{\left( {\frac{q}{m}} \right)B}}\quad \quad ....\left( 1 \right)$$Let $\frac{q}{m} = \alpha $ is called as specific charge
$\therefore$ the equation $(1)$ can be written as$$r = \frac{v}{{\alpha B}}$$$K.E$ of the particles $K=\frac {P^2}{2m}$$$P = \sqrt {2mK} $$ If charged particles is acceleration by potential $V,$ then $$K=qV$$$$\therefore \quad r = \frac{P}{{qB}} = \frac{{\sqrt {2mk} }}{{qB}} = \frac{{\sqrt {2mqV} }}{{qB}}\quad \quad ...\left( 2 \right)$$Time period $$T = \frac{{{\text{length of path}}}}{{{\text{speed}}}}$$$$T = \frac{{2\pi r}}{v} = \frac{{2\pi \times \frac{{mv}}{{qB}}}}{v}$$$$T = \frac{{2\pi m}}{{qB}}$$ Also linear frequency of rotation$$f = \frac{1}{T} = \frac{{qB}}{{2\pi m}}$$ And angular frequency $$\omega = 2\pi f = \frac{{Bq}}{m}$$
NOTE :-
$1.$ Time period $T, f$ and $ω$ are independent of $v.$
$2.$ The velocity at any instant can be written as $$\vec v= v_x \hat i +v_y\hat j$$(ii) When particle is projected out side the field: If the length of the magnetic field is enough, then the angle with which the charged particle emerges out will equal to the angle with which it enters into the field. Thus we have two cases:
$\text {a}.$ Time spend in magnetic field$$t = \frac{T}{2} = \frac{{\pi m}}{{qB}}$$$$PQ = 2r$$
$\text {b}.$ Time spend in magnetic field$$t = \frac{{2\theta }}{{2\pi }}\left[ T \right]$$$$t = \frac{\theta }{\pi }\left[ T \right]$$where$$T = \frac{{2\pi m}}{{qB}}$$$$PQ = 2r\sin \theta $$
Cases 3 :- When particle is projected at an angle $θ,$ which is not equal to $0^\circ, 90^\circ$ or $180^\circ$ :
The velocity of the particle can be resolved into two components; one along $\vec B$ that is $v\cos θ$ and other perpendicular to $\vec B$ that is $v \sin θ.$ The velocity $v \sin θ$ will provide circular path and the velocity $v \cos θ$ will provide a straight line. The resulting path is a helical path.
i. preprocessThe radius of the helical path $$r = \frac{{m{v_ \bot }}}{{qB}} = \frac{{mv\sin \theta }}{{qB}}$$
ii. Since time period of rotation does not depend on velocity, therefore$$T = \frac{{2\pi m}}{{qB}}$$iii. Pitch $(p) :$ The centre to centre distance between two consecutive circular paths is called
pitch. Thus$$p=v \cos \theta \times T$$$$p = v\cos \theta \times \frac{{2\pi m}}{{qB}}$$
$$\therefore \quad p = \frac{{2\pi \;mv\cos \theta }}{{qB}}$$
Comparison of paths in $\vec E-$ field and $\vec B -$ field
