Motion of a particle in a magnetic field and viscous medium

[PumpkinCougar95]

Homework Statement

In a homogeneous, non-magnetic, highly insulating and viscous medium, a moving particle experiences a viscous drag given by the law f→=−bv→. Here b is a positive constant. A particle having charge q is projected with an unknown velocity from a point in the medium. It almost stops after traveling a distance of 10m10m in a straight line. Now a uniform magnetic field is established in the region and the same particle is again launched with the same velocity perpendicular to the magnetic field.

In the presence of above magnetic field, if the particle almost stops at a point 6m from the point of projection, then find the magnitude of magnetic field in terms of q and b.

Homework Equations

The Attempt at a Solution

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https://drive.google.com/file/d/0B2jHGkWhC0E2bElTelB3aVNZTXc/view?usp=sharing
https://drive.google.com/file/d/0B2jHGkWhC0E2ajljRjJqempLeXM/view?usp=sharing
https://drive.google.com/file/d/0B2jHGkWhC0E2T2ZjRzE4eGh3eFk/view?usp=sharing

w is the angular velocity about the origin^. X,Y are the displacement of the particle in X and Y direction when v=0.

The trajectory is a spiral. I was able to find the velocity v as a function of time .
I was able to find the trajectory of the particle in polar coordinates( taking the centre of the spiral as the origin), and the distance of the starting position from the origin. But i can't find the magnetic field. In fact, I think that the magnetic field may not be constant. Am I correct? If not then how to solve this questions?
https://drive.google.com/file/d/0B2jHGkWhC0E2bElTelB3aVNZTXc/view?usp=sharing https://drive.google.com/file/d/0B2jHGkWhC0E2ajljRjJqempLeXM/view?usp=sharing https://drive.google.com/file/d/0B2jHGkWhC0E2T2ZjRzE4eGh3eFk/view?usp=sharing

 

[NFuller]

I'm having some difficulty following your work, some text describing what you are doing would be helpful. I'm not sure if working through the problem in cylindrical coordinates as you have done is any easier than working through it with Cartesian coordinates. Adding vectors in curvilinear coordinates requires you to convert them into Cartesian coordinates anyways.

But i can't find the magnetic field. In fact, I think that the magnetic field may not be constant. Am I correct? If not then how to solve this questions?

I'm pretty sure you are supposed to assume the field is constant, even though it is not explicitly stated in the problem.

I suggest you start by setting up a Cartesian coordinate system where the magnetic field points in the $\hat{z}$ direction $\mathbf{B}=B\hat{z}$ and the initial velocity is in the $\hat{y}$ direction $\mathbf{v}_{0}=v_{0}\hat{y}$. Then using Newton's second law
$$\mathbf{F}=m\dot{\mathbf{v}}=-b\mathbf{v}+q\mathbf{v}\times\mathbf{B}$$
Plugging in $\mathbf{B}=B\hat{z}$ and taking the cross product gives
$$m\dot{\mathbf{v}}=-b\mathbf{v}+qB(v_{y}\hat{x}-v_{x}\hat{y})$$
This leads to the first order linear system
$$m\dot{v}_{x}=-bv_{x}+qBv_{y}$$
$$m\dot{v}_{y}=-bv_{y}-qBv_{x}$$
Do you know how to solve this system?

 

[PumpkinCougar95]

I think this should reduce to a damped oscillation type of equation if you substitute Vy in Vx.

Vy = m(dVx/dt + bVx )/qB

 

[NFuller]

I think this should reduce to a damped oscillation type of equation

Yes, that's how the spiral trajectory occurs. In order to solve the system, you should first write the system as a matrix equation.
$$\dot{\mathbf{v}}=\mathbb{A}\mathbf{v}$$
where
$$\mathbb{A}=\begin{bmatrix}-b/m & qB/m\\ -qB/m & -b/m\end{bmatrix}$$
The solution will be of the form
$$\mathbf{v}=\vec{\chi}e^{\alpha t}$$
Plugging this back into the matrix equation leads to the requirement for eigensystems
$$(\mathbb{A}-\mathbb{I}\alpha)\vec{\chi}=0$$
Where the values of $\alpha$ are found by requiring that
$$\text{Det}(\mathbb{A}-\mathbb{I}\alpha)=0$$
Is this type of math familiar to you?

 

[PumpkinCougar95]

I do know how to solve linear equations with matrices, but that is all.
Anyways, even if you are able to get the trajectory then how would solve for the magnetic field?

Edit: Also, I realized that one of my equations was wrong so don't pay attention to my earlier attempt.

 

[TSny]

Write
$m\dot{v}_{x}=-bv_{x}+qBv_{y}$
$m\dot{v}_{y}=-bv_{y}-qBv_{x}$

as

$m\dot{v}_{x}=-b\dot x+qB \dot y$
$m\dot{v}_{y}=-b\dot y-qB \dot x$

Can you immediately integrate each of these with respect to time?

 

[PumpkinCougar95]

Yes you can, and by integrating,

$m\dot{v}_{x}=-b\dot x+qB \dot y$

$m\dot{v}_{y}=-b\dot y-qB \dot x$

we get
$m{v}_{x}=-b x+qB y$$m{v}_{y}=-b y-qB x$

from which we get:

$$ (Vx)^2 + (Vy)^2 = V^2 = (10 \frac {b}{m} e^{-bt/m})^2$$

we get

$$x^2 + y^2 = \frac { 100b^2 e^{-2bt/m}} { q^2B^2 + b^2}$$

which can be interpreted as a circle with an exponentially decaying radius.
But how to calculate the field?

 

[NFuller]

@TSny's method is more straight forward and gives the velocity as a function of position where my suggestion gives it as a function of time. Since we are really only interested in the final position of the particle, let's follow TSny's suggestion, which is to just integrate both of the equations
$$m\dot{v}_{x}=-b\dot x+qB \dot y$$
$$m\dot{v}_{y}=-b\dot y-qB \dot x$$
with respect to time. This should be a trivial integration so don't overthink it. Also remember to include integration constants.

 

[TSny]

Yes you can, and by using

$$ (Vx)^2 + (Vy)^2 = V^2 = (10 \frac {b}{m} e^{-bt/m})^2$$

we get

$$x^2 + y^2 = \frac { 100b^2 e^{-bt/m}} { q^2B^2 + b^2}$$

which can be interpreted as a circle with an exponentially decaying radius.
But how to calculate the field?

OK, I think your result for $x^2 + y^2$ is correct for appropriately chosen initial conditions, except I think the $100$ needs to be squared. [EDIT: Sorry, your 100 is correct.] What did you take to be the initial conditions for $x, y, v_x$ and $v_y$?

You can find $B$ by considering your result at $t = 0$ and at $t = \infty$.

So, you have essentially solved it. The way I was thinking about it (in line with @NFuller's comments) is to go back and integrate $m\dot{v}_{x}=-b\dot x+qB \dot y$ and $m\dot{v}_{y}=-bv_{y}-qBv_{x}$ to get

$m v_x=-b x + qB y + C_x$
$m v_y=-b y - qB x + C_y$

where $C_x$ and $C_y$ are integration constants. I think it's convenient to choose the initial conditions such that the particle is projected from the origin in the +x direction. The values of $C_x$ and $C_y$ are determined by these initial conditions. The final values of x and y for the particle can be found from the above equations by letting $v_x = v_y = 0$. Then you can solve for $B$ by knowing that the final position is 60 6m from the initial position. You don't need to actually find $x$ and $y$ as functions of time.

 

[PumpkinCougar95]

Ok , So by considering your initial conditions we have

$mVy = -by -qBx$ which implies at $Vy=0$ $y=-\frac{qBx}{b}$
and
$mVx=-bx + qBy + 10b$

now since $sqrt(x^2 + y^2) = 6$

I was able to find $B= \frac {4b}{3q}$ which matches the answer.

Thanks for you Help!

 

[TSny]

OK, good.