The uniform magnetic field $B$ does not apply any force on the charged particle (say, electron) in the parallel direction that is $F_{\parallel}=q\,v_{\parallel}\,B\sin 0=0$. MECHANICS Since the force acts perpendicular to its velocity, the force does not do any work. So, the magnetic force also provides the centripetal force to the charge. The difference is that a moving charge has both electric and magnetic fields but a stationary charge has only electric field. A particle of charge q and mass m moves in XY plane. Helical path is the path of the motion of a charged particle when enters at an angle in a uniform magnetic field . A charged particle moving along positive x-direction with a velocity v enters a region where there is a uniform magnetic field B = − B k ^,from x=0 to x=d.The particle gets deflected at an angle θ from its initial path.The specific charge of the particle is Let us consider a uniform magnetic field of induction B acting along the Z-axis. Particle with mass m, charge in a uniformed magnetic field Bz. The force F acting towards the point O acts as the centripetal force and makes the particle to move along a circular path. And you got, \[f = \frac{|q|B}{2\pi \, m} \tag{5} \label{5}\]. This magnetic lorentz force provides the necessary centripetal force. Applying Newton's second law of motion and balancing the centripetal force with the magnetic force we get a formula for radius of helical path as \begin{align*} F&=m\,a_r\\q\,v_{\bot}\,B&=m\,\frac{v_{\bot}^{2}}R\\\Rightarrow R&=\frac{m\,v_{\bot}}{q\,B}\\&=\frac{mv\,\sin\theta}{qB}\end{align*} Where $m$ is the mass of the charged particle. TERMS AND PRIVACY POLICY, © 2017-2020 PHYSICS KEY ALL RIGHTS RESERVED. The magnitude of magnetic force on the charge (if you haven't read this article about magnetic force, review that article) is, \[F =|q|vB\sin \theta = qvB \tag{1} \label{1}\], where $\theta$ is the angle between $\vec v$ and $\vec B$ but the angle is always a right angle, so $\sin \theta = 1$. Let's see what happens next. Those two above motions, uniform motion parallel to the field $B$ and uniform circular motion perpendicular to the field $B$, creates the actual path of a charged particle in a uniform magnetic field $B$ which is similar to a spring and is called a spiral or helical path. The difference is that a moving charge has both electric and magnetic fields but a stationary charge has only electric field. So, what we got here is an expression for the radius of the circle in which the charge moves under the action of magnetic force. Physics problems and solutions aimed for high school and college students are provided. © 2015 All rights reserved. Copyright © 2018-2021; All Rights Reserved. Since the magnetic force is directed perpendicular to the plain containing $\vec v$ and $\vec B$, that is the magnetic force $\vec F$ is always perpendicular to $\vec v$, the charge moves in a circle of arbitrary radius $r$ (see fig). You can easily understand the proportionality of the radius to other related quantities from the above equation. Let us consider a uniform magnetic field of induction B acting along the Z-axis. In the parallel case, there is no force on the particle but in the perpendicular one, there is a centripetal acceleration toward the center. But if you consider a particular instant of motion, it has a velocity vector $\vec v$. This is the main factor that creates a spiral or helical path. The angular speed $\omega$ is related to the linear speed $v$ and radius $r$, that is $\omega = v/r$, so the angular speed using Equation \eqref{3} is, \[\omega = \frac{|q|B}{m} \tag{4} \label{4}\], You know that the frequency $f$ of the rotation is $\omega / 2\pi$. This java applet show charged particle motion in uniform magnetic field. Let us consider a uniform magnetic field of induction B acting along the Z-axis. For example you can hold ionized gas of very high temperature such as $10^6 \text{K}$ in a magnetic bottle which can destroy any material if comes in contact with such a high temperature. I considered the charge is moving with speed $v$ not with velocity $\vec v$ because the velocity changes continuously, that is the charge's direction is changing continuously. In addition, there are hundreds of problems with detailed solutions on various physics topics. (2D case) When the charged particle is within a magnetic field, the radius of the circular motion is quite small and the frequency is huge. $\begingroup$ Related : Motion of charged particle in uniform magnetic field and a radially symmetric electric field. $\endgroup$ – Frobenius Nov 9 at 1:06 1 $\begingroup$ The solution is a helix-- … If the charge is negative the rotation is clockwise. But if the angle is not a right angle there is also a component of velocity vector parallel to the magnetic field. Deviation of charged particle in uniform magnetic field: Case 1: Suppose a charged particle enters perpendicular to the uniform magnetic field if the magnetic field extends to a distance ‘x’ which is less than or equal to radius of the path.


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