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# Charged Particle in Coexisting Electric and Magnetic Fields

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This post is categorized as ‘Lecture’ because, even if it is based on a single problem, some important concepts of Physics are illustrated in this.

Question:

A uniform electric field of strength $E$ acts parallel to the positive Y-axis. A uniform magnetic field of strength $B$ acts parallel to the positive Z-axis. A particle of charge $q$ with mass $m$ placed at the origin O is released from rest. Find out the following:

1. the position vector of the particle as a function of time
2. if there will be any double point in the trajectory curve; if yes, then will it be a node or a cusp
3. the shortest time interval between points where the particle comes to rest
4. the displacement of the particle along the Y-axis when the velocity of the particle becomes parallel to the X-axis for the first time
5. the velocity of the particle when it becomes parallel to X-axis
6. if the particle will have a velocity purely along the Y-axis

If you prefer watching over reading, you can also watch the video lecture below.

This is a key problem for students preparing for JEE Advanced, JEE Main, and NEET (undergraduate). Many questions of varying standards and toughness can be formed on its basis, some of which have been put here in the question statement. For example, the magnetic field working perpendicular to the velocity vector all the time is not going to produce any work. Thus the question can be a little modified to be easily solved by using the Work-Energy principle so that the work done by the electric field can be equated to the kinetic energy of the particle. However, in this particular problem, that technique is useless. That’s why we must understand the Physics and Mathematics involved in this question. A short tutorial post explaining how a sample variant of this question can be asked has also been published.

In the diagram below, the distribution of the electric field $\vec{E}$ and the magnetic field $\vec{B}$ is shown along the Y and Z axes respectively. The particle of charge $q$ and mass $m$ is placed at the origin O. Starting from the time $t=0$, the path of the particle at any instant $t$ is shown along with its velocity, tangential to the path, making an angle $\theta$ with the Y-axis. The corresponding $x$ and $y$ components of the velocity, the Coulomb force $\vec{F_c}$ (electric force), and the Lorentz force $\vec{F_l}$ (magnetic force) along with its components are also shown.

Diagram 1:

It is clear that:

The Coulomb force, $\vec{F_c}=qE\:\hat{j}$

The Lorentz force $\vec{F_l}=q{\vec{v}}\times \vec{B}$

Magnitude of $\vec{F_c}=F_c=qvB\:\:\:(\because \vec{v}\perp\vec{B})$

If at any time $t$, the velocity vector $\vec{v}$ of the particle makes an angle $\theta$ with the Y-axis, then:

$v_x=v sin \theta$ & $v_y=v cos \theta$

Net force along X, $F_x=F_{l_x}=F_lcos\theta=qBvcos\theta=qBv_y$

Net force along Y, $F_y=F_c-F_{l_y}=F_c-F_lsin\theta=qE-qB v sin\theta=qE-qBv_x$

Now, let’s try to understand what really happens when the particle at the origin O is released from rest. This is shown in the diagram below.

Diagram 2:

At the time $t=0$, the particle is at O and is at rest. At this instant, there is only one force acting on it. This is the Coulomb force, $\vec{F_c}$ along the +Y direction. From this, the acceleration along the Y-axis, $a_y$ can be found as:

$F_c=qE\implies a_y=\frac{qE}{m}$

This vertically upward acceleration will cause the particle to acquire a differential vertical velocity $dv_{y_1}$ in a differential time interval of $dt_1$. The only possible force along X is the Lorentz force ($\vec{F_l}$). Since $\vec{F_l}$ depends on the net velocity ($\vec{v}=v_x\:\hat{i}+v_y\:\hat{j}$) of the particle, till $t$ reaches the value of $dt_1$, $F_l$ remains negligibly 0. Thus far, the particle moves mostly along Y. In the next time interval $dt_2$, at $t=dt_1 + dt_2$, the vertical velocity $dv_{y_2}$ of the particle causes a horizontal differential Lorentz force ($dF_l$), thus, a horizontal differential acceleration ($da_x$) and hence a horizontal differential velocity ($dv_{x_2}$). At this instant, the particle begins to deviate from moving purely along Y into a curve in the XY plane or Z plane. Let’s summarize all the data we have for the initial moment, which can also be called boundary conditions, i.e., at $t=0$

Now, we are ready to answer the first question, i.e., the position vector $\vec{r}$ of the particle as a function of time:

$$\boxed{\vec{r}=x(t)\:\hat{i}+y(t)\:\hat{j}}$$

To find $x(t)$ and $y(t)$, we have to begin with $a_x$ and $a_y$ at any time $t$ where the velocity of the particle $\vec{v}$ makes an angle $\theta$ with the +Y and proceed as follows:

$a_x=\frac{F_l cos\theta}{m}=\frac{qB}{m}v_y$ $\:\:\:Eq^n\:(1)$

$a_y=\frac{F_c}{m}-\frac{F_l sin\theta}{m}=\frac{qE}{m}-\frac{qB}{m}v_x$ $\:\:\:Eq^n\:(2)$

Now, we can clearly see that $a_x$ depends on $v_y$ and $a_y$ depends on $v_x$. So, these two quantities are entangled. In order to solve this situation, we are going to use a few mathematical techniques. Let’s differentiate $a_y$ once WRT $t$ and get rid of the constant term. This will also differentiate $v_x$ and give us $a_x$ which depends only on $v_y$. Thus, we can have an equation purely along Y and then we’ll see what’s next. Okay? Let’s do that.

$\frac{d}{dt}(a_y)=-\frac{qB}{m}a_x=-(\frac{qB}{m})^2v_y$ $\:\:\:Eq^n\:(3)$

$\because a_y={v_y}^{‘} \implies \frac{d}{dt}(a_y)={v_y}^{”}$

Thus $Eq^n\:(3)$ becomes: ${v_y}^{”}=-(\frac{qB}{m})^2v_y$ $\:\:\:Eq^n\:(4)$

One way of solving $Eq^n\:(4)$, is to use the technique of solving the simple harmonic motion equation.

If we get the equation

$r^{”}=-\omega ^2 r$

then the general solution is

$r=r_{max} sin (\omega t+\delta)$

in which we can use boundary conditions to obtain the particular solution.

For our case, the general solution is

$v_y=v_{y_{max}} sin\: (\omega t+\delta)$ where $\omega=\frac{qB}{m}$

From the boundary conditions, we know that, at $t=0$, $v_y=0$ and $v_y^{‘}=\frac{qE}{m}$.

This gives us the following:

$0=v_{y_{max}} sin\:\delta \implies \delta=0\implies v_y=v_{y_{max}} sin\: \omega t$

$\frac{qE}{m}=\omega v_{y_{max}}\implies v_{y_{max}}=\frac{E}{B}$

Thus, the particular solution we need and now we have is:

$$\boxed{v_y=\frac{E}{B}\:sin\:\omega t}$$ $\:\:\:Eq^n\:(5)$

The other and mathematically more rigorous way of solving $Eq^n\:(4)$, is to use the standard method of solving linear homogeneous differential equations of second order:

For an equation of the form $r^{”}+ar^{‘}+br=0$ where $a$ and $b$ are constants, the quadratic characteristic equation is:

$\lambda^2+a\lambda+b=0$

Then we need to find the roots of this equation and proceed to the general solution depending upon whether the roots are (i) real and distinct, (ii) real and repeated or (iii) complex conjugates. In our case, if we rewrite our $Eq^n\:(4)$ as:

${v_y}^{”}+(\frac{qB}{m})^2v_y=0$,

we shall see that, $a=0$ and $b=(\frac{qB}{m})^2$.

In this form, we can see that the quadratic characteristic equation is going to give us imaginary roots $\pm i\omega$, where $\omega=\frac{qB}{m}$ and $i=\sqrt {-1}$.

Thus, we can obtain the general solution:

$v_y=e^{\frac{-a}{2}t}[C_1\:cos\:(\omega t) + C_2\:sin\:(\omega t)]$ (where $C_1$ and $C_2$ are arbitrary constants)

From the boundary conditions, we know that, at $t=0$, $v_y=0$ and $v_y^{‘}=\frac{qE}{m}$.

Thus, our general solution reduces to the particular solution:

$\boxed{v_y=\frac{E}{B}\:sin\:\omega t}$ $\:\:\:Eq^n\:(5)$ (as obtained above)

Now that we have $v_y$, we can write $a_x$ as:

$a_x=\frac{qB}{m}v_y=\frac{qE}{m}\:sin\:\omega t$

Since $a_x=\frac{dv_x}{dt}\implies dv_x=\frac{qE}{m}\:sin\:\omega t\:dt$

By integrating the above equation for $v_x$ varying with limits from $0$ to $v_x$ and time from $0$ to $t$, we can get $v_x$.

$\int_{0}^{v_x} dv_x=\frac{qE}{m}\int_{0}^{t} sin\:\omega t\:dt$

$$\implies\boxed{v_x=\frac{E}{B}(1-cos\:\omega t)}$$ $\:\:\:Eq^n\:(6)$

Now, we can find both $x(t)$ and $y(t)$ by integrating $\:\:\:Eq^n\:(6)$ and $\:\:\:Eq^n\:(5)$ respectively, within proper limits.

$v_x=\frac{E}{B}(1-cos\:\omega t)$

$\implies \frac{dx}{dt}=\frac{E}{B}(1-cos\:\omega t)$

$\implies dx=\frac{E}{B}(1-cos\:\omega t)dt$

$\implies \int_{0}^{x} dx=\frac{E}{B}\int_{0}^{t}(1-cos\:\omega t)dt$

$$\implies \boxed{x=\frac{E}{B}(t-\frac{sin\:\omega t}{\omega})}$$ $\:\:\:Eq^n\:(7)$

Similarly $y$ can be found:

$v_y=\frac{E}{B}\:sin\:\omega t$

$\implies \frac{dy}{dt}=\frac{E}{B}\:sin\:\omega t$

$\implies dy=\frac{E}{B}\:sin\:\omega t\:dt$

$\implies \int_{0}^{y} dy=\frac{E}{B}\int_{0}^{t}\:sin\:\omega t\:dt$

$$\implies \boxed{y=\frac{mE}{qB^2}(1-cos\:\omega t)}$$ $\:\:\:Eq^n\:(8)$

Now, using $\:\:\:Eq^n\:(7)$ and $\:\:\:Eq^n\:(8)$, we can answer

Part 1: the position vector as a function of time:

$$\boxed{\vec{r(t)}=\frac{E}{B}(t-\frac{sin\:\omega t}{\omega})\:\hat{i}+\frac{mE}{qB^2}(1-cos\:\omega t)\:\hat{j}}$$

Part 2: if there will be any double point in the trajectory curve; if yes, then will it be a node or a cusp

To answer this question, we need to trace the curve of the particle’s trajectory and see if there are any double points on it, i.e., points where the curve crosses itself with real and distinct tangents to the branches of the curve at that point (node) or touches itself with a single real tangent to all the branches at that point (cusp).

Curve Tracing is an interesting topic in calculus and ThePhysicist will cover that topic through dedicated posts. But here we can find the answer through some simple analysis of $x(t)$, $y(t)$, $v_x (t)$, and $v_y (t)$.

$x(t)$ contains a linear part as $t$ and a negative periodic part as $\frac{sin\:\omega t}{\omega}$. Thus $x(t)$ is going to increase with time without bounds but the rate of increase will fluctuate periodically. This can be seen in $v_x$ as it is also periodic, being $0$ (minimum) when $t=\frac{2n\pi}{\omega}$ and $\frac{2E}{B}$ (maximum) at $t=\frac{(2n+1)\pi}{\omega}$, for $n=0,1,2,3,…$.

$y(t)$ is a purely periodic function being $0$ (minimum) when $t=\frac{2n\pi}{\omega}$ and $\frac{2mE}{qB^2}$ (maximum) at $t=\frac{(2n+1)\pi}{\omega}$, for $n=0,1,2,3,…$. This can be seen in $v_y$ as it is a sine function, being $0$ (mean) when $t=\frac{n\pi}{\omega}$, $\frac{E}{B}$ (maximum) at $t=\frac{\pi}{\omega}(\frac{4n+1}{2})$, $-\frac{E}{B}$ (minimum) at $t=\frac{\pi}{\omega}(\frac{4n+3}{2})$ for $n=0,1,2,3,…$.

Thus, the curve starts from O, then moves continuously along +X, while rising from $0$ in Y to $\frac{2mE}{qB^2}$ and then back to $0$. This can help us plot a qualitative cycle of the curve as shown below.

Diagram 3:

If we wish to verify what we have concluded about the shape of the trajectory, we can do it very easily using some easily available CAS tools. There are some which have apps for Android, iOS, Mac OS and Windows. Some also have web apps. The most easily available web apps are desmos and GeoGebra.

For simplicity for plotting the trajectory in desmos, let’s normalize all the parameters, $E$, $B$, $q$, $m$, and $\omega$ to unity. This will give us a parametric form of the trajectory $\left(\left(t-\sin t\right),\left(1-\cos t\right)\right)$ which we can put in desmos and obtain the trajectory as below.

Diagram 4:

From this, we can say that, starting from O, at every $\frac{2\pi E}{\omega B}$ units along +X, we shall have a cusp on the trajectory.

Part 3: the shortest time interval between points where the particle comes to rest

The particle comes to rest when $v_x=v_y=0$. This happens when $sin\:\omega t=0$ and $cos\:\omega t=1$. The minimum period for that is $\omega t=2\pi$. Hence, the minimum time interval between two such consecutive points is

$$\boxed{\delta t=\frac{2\pi}{\omega}}$$.

Part 4: the displacement of the particle along the Y-axis when the velocity of the particle becomes parallel to the X-axis for the first time

The velocity of the particle becomes parallel to the X-axis when $v_y=0$ but $v_x$ is maximum. This happens for the first time when $\omega t=\pi\implies t=\frac{\pi}{\omega}$.

Putting that value of $t$ in $\:\:\:Eq^n\:(8)$, we get the required answer.

$$\boxed{y=\frac{2mE}{qB^2}}$$

Part 5: the velocity of the particle when it becomes parallel to X-axis

This can be found very simply as it is related to the previous question. When the velocity of the particle becomes parallel to the X-axis for the first time, $t=\frac{\pi}{\omega}$ for which $\:\:\:Eq^n\:(6)$ shows that $$\boxed{v_x=\frac{2E}{B}}$$.

Part 6: if the particle will have a velocity purely along the Y-axis

This is a question based on the sense of trigonometry. It asks if ever, $v_x=0$ would happen when $v_y\neq0$. $\:\:\:Eq^n\:(6)$ shows that $v_x=0\implies\omega t=2n\pi$ for which $\:\:\:Eq^n\:(6)$ shows that $v_y=0$.

Hence, there never will be a point on the trajectory where the velocity is purely along the Y-axis.

The differential approach illustrated in diagram 2 is only for illustration purpose.

If you prefer watching a lecture over reading, you can watch the video lecture below.

This concludes the lecture. If you have any requests, suggestions, or complaints regarding this lecture, please post your comment below. For new questions, please use the General Physics Forum and I will respond there with the help you require.

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