Where does the particle stop?

In the first case, when the particle moves in a straight line, by Newton's second law, the following equality holds: $m \frac{\mathrm d v}{\mathrm d t} = -kv \rightarrow m \frac{\mathrm d v}{\mathrm d t} = -k \frac{\mathrm d s}{\mathrm d t}$ $\rightarrow m \int \,\mathrm{d}v = -k \int \,\mathrm{d}s \rightarrow mv_{0} = ks \rightarrow \frac {mv_{0}}{k} = s = 0.10 \,\ m$

When the particle moves in a magnetic field, it is under the effect of 2 forces: the drag force with magnitude $kv$ which is opposite to the velocity and the magnetic force $Bqv$ which is perpendicular to the velocity. Drawing out the diagram of the particle with the velocity and the forces given, we can see that the net force on the particle always forms with the velocity vector an angle $\alpha = \pi - \theta$ , where $\sin \theta = sin \alpha = \frac {Bq}{\sqrt{B^2 q^2 + k^2}}$ .

Since the particle spirals and comes to a stop, this implies that the net force always points to the final position (otherwise it will not be drawn toward that point) (in fact, the particle's trajectory is a logarithmic spiral), which means that the acceleration always point toward the final position. This statement implies that if we draw a line (which is called the radius) from the final position to any point on the trajectory, this line always contain the net force. Hence, the angle between the radius and the velocity is equal to the angle between the net force and the velocity, which is $\alpha$ .

The particle is moving in a magnetic field, therefore its angular velocity is given by $\omega = \frac {Bq}{m}$ .

In the second case, the particle starts with velocity $v_{0}$ and undergoes circular motion with radius is $L_{1}$ . We have: $v_{0} \sin \theta = \omega L_{1} \rightarrow L_{1} = \frac {v_{0} \sin \theta}{\omega} = \frac {mv_{0} \sin \theta}{Bq} = \frac {mv_{0}}{\sqrt{B^2 q^2 + k^2}} = \frac {\frac{mv_{0}}{k}}{\sqrt{\frac {B^2 q^2}{k^2} + 1}}$

Since $L_{1} = 0.06$ and $\frac {mv_{0}}{k} = 0.10$ , we get $\frac {Bq}{k} = \frac {4}{3}$ .

In the third case, the magnetic force is $\frac {Bqv}{2}$ . Similarly, the angle between the velocity and the net force is $\alpha' = \pi - \theta'$ , where $\sin \theta' = \sin \alpha' = \frac {Bq}{\sqrt{B^2 q^2 + 4k^2}}$ . Also $\omega' = \frac {Bq}{2m}$ .

Similarly, $v_{0} \sin \theta' = \omega' L_{2}$ , which gives $L_{2} = \frac {\frac{2mv_{0}}{k}}{\sqrt{\frac {B^2 q^2}{k^2} + 4}} = \frac {2*0.10}{\sqrt {\frac {16}{9} + 4}} = 0.0832 \,\ (m)$ .

Using the first piece of info, we have, $m\times a=-kv$ which can be written as, $m(dv/dx)v= -kv$ which simplifies to, $m(dv/dx)= -k$ . Integrating this and putting the limits(from velocity v to 0), we get, mv=kl Moving on to the real part of the question, lets analyse in x and y directions, let the velocities be v(x) and v(y). Writing the force equation gives, $m(d(v(x))/dt)= -(qB)v(y) - kv(x)$ , and, $m(d(v(y))/dt)= (qB)v(x) - kv(y)$ . Here, I have also considered the magnetic force which is $q(v \times B)$ . Integrating these two within the limits, we get an equation involving the x and y coordinate, v(x) and v(y). We are almost there. When it comes to a stop, then v(x)=v(y)=0, so solving the linear equation we get x, y, and thus the displacement $\sqrt{x^2 +y^2}$ . The final value comes out to be of the form, $d = mv/((q.b)^2 +k^2)^{1/2}) = kl/(((qb)^2 +k^2)^{1/2})$ (Using the first piece of info). Now putting the values given, we get $qB = (4/3)k$ . If B is halved, then d becomes $3/\sqrt{13} \times l$ which is the answer.

\­(v {0}\­), q, B are the velocity, charge and magnetic field. We have the equation of the charge's motion: \­( m\frac{d v }{dt}=q v \land B -k v \­) Then \­( md v =qd r \land B -kd r \­) After make an integral for two members from the initial time to the stop, we have: \­( m v 0 }=q r \land B -k r \­) Therefore: \­( L=\frac{mv {0}}{\sqrt{q^2B^2+k^2}}\­) With L is the displacement of the charge. Finally: \­( L {2}=\frac{2L 0L 1}{\sqrt{L^2 03L^2 1\­) Numerical application: \­( L_2\­)=0.083 (m)

NOTE: This solution uses a bit of calculus, I couldn't avoid it.

Let us consider the first case.
Applying Newton's Law,
$\displaystyle \vec{F_{net}} = -k\vec{v}$

Since this is a 1-Dimensional case, we drop the vector notation and focus on only the magnitudes.

$\displaystyle m \frac{dv}{dt} = -kv$

$\displaystyle m \frac{dv}{dx} \frac{dx}{dt} = -kv$

$\displaystyle m \frac{dv}{dx} v = -kv$

$\displaystyle m\cdot dv = -k\cdot dx$

$\displaystyle m\int\limits_{v_o}^0 dv = -k\int\limits_0^{L_o} dx$

$\displaystyle \boxed{mv_o = kL_o}$

Also note that, $v = v_o e^{(-k/m)t}$

Now, for the second case,
We shall go through a logical thought process.
We can agree on the fact that the second case is the combination of the first case and a circular motion.

So, we shall try combining the two concepts.

$\bullet$ For the first case, the solution was, $v = v_o e^{(-k/m)t}$

$\bullet$ For a circular motion the solution is,
$\displaystyle v_x = v_o \sin(wt)$
$\displaystyle v_y = v_o \cos(wt)$ , where $\displaystyle w = \frac{Bq}{m}$

So, combining the two concepts, we have the possible (also logical) solution as,

$\displaystyle v_x = v_o e^{-\lambda t} \sin(wt)$
$\displaystyle v_y = v_o e^{-\lambda t} \cos(wt)$ , where $\displaystyle \lambda = \frac{k}{m}$

NOTE: You can check this solution by forming a differential equation for this case, and plugging in this solution.

Now, to calculate the displacement,
We calculate the final $\displaystyle (x,y)$ coordinates of the particle, by integrating the above two equations, just as with the first case.

$\displaystyle x = \int\limits_0^{\infty} v_o e^{-\lambda t} \sin(wt) dt$ and,

$\displaystyle y = \int\limits_0^{\infty} v_o e^{-\lambda t} \cos(wt) dt$

We finally get,

$\displaystyle x = v_o\frac{w}{w^2+\lambda^2}$

$\displaystyle y = v_o\frac{\lambda}{w^2+\lambda^2}$

Thus, the displacement is,

$\displaystyle \boxed{ L_1 = \sqrt{x^2+y^2} = \frac{v_o}{\sqrt{w^2+\lambda^2}} = \frac{mv_o}{\sqrt{k^2+B^2q^2}}}$

Since the magnetic field is halved , we have,

$\displaystyle L_2 = \frac{mv_o}{\sqrt{k^2+\frac{B^2q^2}{4}}}$

Using all the boxed equations and the values given ,

$\displaystyle L_2 = \frac{3}{\sqrt{13}} (0.1)m \approx \boxed{0.08320m}$

$Let\quad the\quad particle\quad initially\quad travel\quad along\quad x-axis\quad \\ starting\quad from\quad origin.\\ At\quad any\quad instant,\\ { F }_{ x }=-qB{ V }_{ y }-k{ V }_{ x }\\ { F }_{ y }=\quad qB{ V }_{ x }-k{ V }_{ y }\\ The\quad above\quad equations\quad can\quad be\quad written\quad as\\ m\frac { { dV }_{ x } }{ { dt } } =-qB\frac { dy }{ dt } -k\frac { { dx } }{ dt } \\ m\frac { { dV }_{ y } }{ { dt } } =\quad qB\frac { dx }{ dt } -k\frac { { dy } }{ dt } \\ Cancelling\quad out\quad dt\\ m{ dV }_{ x }=-qBdy-kdx\\ m{ dV }_{ y }=\quad qBdx-kdy\\ Let\quad x\quad and\quad y\quad be\quad the\quad final\quad displacement\quad \\ along\quad x\quad and\quad y\quad axis\quad respectively.\\ \displaystyle m\int _{ { V }_{ o } }^{ 0 }{ { dV }_{ x } } =-qB\int _{ 0 }^{ y }{ dy } -k\int _{ 0 }^{ x }{ dx } \\ \displaystyle m\int _{ 0 }^{ 0 }{ { dV }_{ y } } =\quad qB\int _{ 0 }^{ x }{ dx } -k\int _{ 0 }^{ y }{ dy } \\ Evaluating\quad the\quad integral,\quad we\quad get\\ -m{ V }_{ o }=-qB(y)-k(x)\\ \quad \quad \quad 0=\quad qB(x)-k(y)\\ Solving\quad the\quad above\quad Linear\quad equations,\quad we\quad get\\ x=\frac { m{ V }_{ o }k }{ { (qB) }^{ 2 }+{ k }^{ 2 } } \quad and\quad y=\frac { m{ V }_{ o }qB }{ { (qB) }^{ 2 }+{ k }^{ 2 } } \\ s=\sqrt { { x }^{ 2 }+{ y }^{ 2 } } \\ \boxed { s=\frac { m{ V }_{ o } }{ \sqrt { { (qB) }^{ 2 }+{ k }^{ 2 } } } } \\ Case\quad 1:\quad B=0,s=0.1m\\ 0.1=\frac { m{ V }_{ o } }{ k } \\ Since,\quad m\quad and\quad { V }_{ o }\quad are\quad arbitrary,\quad \\ we\quad might\quad as\quad well\quad take\quad m=1kg\quad and\quad { V }_{ o }=1m/s\\ Which\quad gives\quad us\quad k=10.\\ Case\quad 2:\quad B=B_{ o },s=0.06\\ Plugging\quad in\quad the\quad values,\quad we\quad get\quad { qB_{ o } }=\frac { 40 }{ 3 } .\\ Case\quad 3:\quad B=\frac { B_{ o } }{ 2 } ,s=L_{ 3 }\\ L_{ 3 }=\frac { 1 }{ \sqrt { { (\frac { qB_{ o } }{ 2 } ) }^{ 2 }+{ k }^{ 2 } } } =\frac { 1 }{ \sqrt { { (\frac { 400 }{ 9 } ) }+100 } } =\boxed { \boxed { 0.0832m } }$