# Magnetic Fields Proof

In the figure below a beam of electrons whose kinetic energy $K$ emerges from a thin-foil “window” at the end of an accelerator tube. A metal plate at distance $d$ from this window is perpendicular to the direction of the emerging beam. I am going to show that the beam can be prevented from hitting the plate if a uniform magnetic field $B>\sqrt{\frac{2mK}{e^{2}d^{2}}}$ is applied.

I start by stating that the Kinetic Energy is given by the formula $K=\frac{1}{2}mv^2$. Because the beam of particles exist within a magnetic field, they experience a $F=ma$, directed radially inward.. Because the applied magnetic field causes them to follow a circular path, $a=\frac{v^2}{r}$ and that magnetic force is $F=qvB$. If we plug all of these values into the radius of a circle that is created when the beam of electrons move perpendicular to the applied uniform magnetic field I obtain the formula $r=\frac{mv}{|q|B}$. If I solve the Kinetic Energy formula for $v$ and then plug it into the circle equation I obtain $r=\frac{m\sqrt{\frac{2K}{m}}}{qB}$. Next, if I solve for $(r=d)$  after squaring both sides and state that the distance of the circle must be less than $d$, I obtain the forumla $d < \sqrt{\frac{m2K}{e^{2}B^{2}}}$. If I now take $\frac{1}{B}$ out of the fraction and multiply each side of the equation by its reciprocal, and then bring $\frac{1}{d}$ under the radical, I obtain the following solution which is the end that I seek, namely $B>\sqrt{\frac{2mK}{e^{2}d^{2}}}$. After using the Right-Hand Rule, I find that the Magnetic Field is pointing out of the page.

The figure below shows a rod of resistive material. Its resistance per unit length increases in the positive direction of the x axis. At any position $x$ along the rod, the resistance $dR$ of a narrow section of width $dx$ is given by $dR=5x dx$, where $dR$ is in ohms and $x$ is in meters. I want to slice off a length of the rod between $x=0$ and some position $x=L$ and then connect that length to a battery with a potential difference $V=5V$. I want the current in the lenght to transfer energy to thermal energy at the rate of $200 W$. I am going to show how to find the position $x=L$ such that these conditions are met.

Since I am dealing with a resistor, I can relate the given rate of energy transfer to thermal energy, $P$, to the Resistance, $R$, using the equation $P=\frac{V^2}{R}$, which is the resistive dissipation of the resistive battery. Essentially electric potential energy is converted to internal thermal energy via collisions between charge carriers and atoms. After plugging in the given voltage for the battery and the rate of thermal energy transfer, and then solving for $R$, I obtain $R=1.25\times10^{-1}\Omega$, which is the resistance of the battery. Next, I can integrate the given differential, $dR=5x dx$, which will equal $R$ and set it equal to $R=1.25\times10^{-1}\Omega$. I am doing this because the rate of thermal energy transfer depends upon the Resistance, which in turn depends on the length of the rod. After applying the Fundamental Theorem of Calculus to the integral and solving for $R$, I obtain $L=2.236\times10^{-1}m$

# Stored Electrical Potential Energy Proof

Let’s imagine a cylindrical capacitor that has radii a and b as in the figure below. I am going to show that half the stored electric potential energy lies within a cylinder whose radius is $r =\sqrt{ab}$.

In this figure the dotted line is the imagined Gaussian surface with length $L$ and radius $R$, such that $a. I need to create a Gaussian surface so that I can apply Gauss’ Laws. On the surface of the Gaussian surface the charge density is $\mu = \frac{1}{2}\epsilon_{0}E$ where $E$ is the Electric Field at points on the Gaussian Surface. Now if I take the formula for the electric field at any point due to an infinite line of charge with uniform linear charge density $\lambda$, namely the formula $E=\frac{\lambda}{2\pi\epsilon_{0}R}$, and plug it into the equation for the charge density on the Gaussian surface, I obtain $\mu = \frac{1}{2}\epsilon_{0}(\frac{q}{2\pi\epsilon_{0}RL})^{2}$, where $\lambda=\frac{q}{L}$. Now I will leave that formula for a moment to develop a volume integral that will sweep out the energy between the cylinders.

I will begin by stating that the charge density per unit volume is $\mu = \frac{\mu}{V}$. This implies that we must think of $\mu$ as a function of the volume which implies that $u(r) = \int_{a}^{R} \mu dV$. This volume integral will give us the energy in the gaussian cylinder. Now I must find a substitution for the volume because the radius will change as I sweep out energy. This will be the volume of a cylinder which is $V=\pi R^{2}L$. But, I need this in differential notation because of the changing radius, which gives me $dV=2 \pi RL dR$. This substitution is important because now the variable of integration and its respective differential are apart of the same volume integral that I will be using. This will become obvious after the next step. Now, I take the energy density formula and the differential volume formula and plug them into the volume integral that I set up to obtain $u(r) = \int_{a}^{R}\frac{q^{2}2\pi RL}{8{\pi^2}\epsilon_{0}{R^2}{L^2}}dR$. After reducing this integral and factoring out the constants, I obtain $u(r)= \frac{q^2}{4\pi\epsilon_{0}L}\int_{a}^{R}\frac{dR}{R}$. After applying the fundamental theorem of Calculus, I obtain  the formula $u(r) = \frac{q^2}{4\pi\epsilon_{0}L}ln\frac{R}{a}$. This is not the total energy for the cylindrical capacitor and is only the energy within the radius of the Gaussian surface. The total energy for the cylinder is the same as this integral, except the limits of integration are different. This integral for the total energy is given by $u(r) = \int_{a}^{b}\frac{q^2dR}{4\pi\epsilon_{0}LR}$, which gives me $\frac{q^2}{4\pi\epsilon_{0}L}ln\frac{b}{a}$. What I now have to show is that half of the total stored electrical potential energy exists within a Gaussian radius that is exactly half of the  actual radius of the cylinder. This implies that $\frac{ \mu_{R}}{\mu_{b}}=\frac{1}{2}$. This now implies that I need to divide the formula for the total stored electrical potential energy into the formula for the stored electric potential energy within the Gaussian radius. This will look like $\frac{ln\frac{R}{a}}{ln\frac{b}{a}} =\frac{1}{2}$. If I solve for R, I obtain the final solution which is $R =\sqrt{ab}$. The proof is complete and now no further steps are necessary.

# Torque problem

Below is a figure of a long, nonconducting, massless rod of length $L$, pivoted at its center and balanced with a block of weight $W$ at a distance $x$ from the left end. At the left and right ends of the rod are attached small conducting spheres with a positive charges $q$ and $2q$, respectively. A distance $h$ directly beneath each of these spheres is a fixed sphere with a positive charge $Q$. I am going to show how to find the distance $x$ when the rod is horizontal and balanced and the value $h$ should have so that the rod exerts no vertical force on the bearing when the rod is horizontal and balanced.

In order for the Rod to be horizontal and balanced, it must be in equilibrium. Because 2 electrostatic forces and 1 weight force cause torque about the axis, the sum of these three torques must add to equal 0 which is notated as $\sum\Upsilon = 0$

The equation for the electromagnetic force, $F_1$, is $\frac{qQ}{4\pi \epsilon_0 h^2}$, and the equation for the electromagnetic force, $F_2$ is $\frac{2qQ}{4\pi\epsilon_0 h^2}$, where $q$ and $Q$ represent respective charges, $h$ is the distance between spheres, and $\epsilon_0$ is the permittivity constant which is a measure of the resistance that is encountered when forming an electric field in a medium.

Now after adding up the torques and setting them equal to zero I obtain this equation: $(\frac{qQ}{4\pi\epsilon_0 h^2})(\frac{L}{2}) + W(x-\frac{L}{2}) - (\frac{2qQ}{4\pi\epsilon_0 h^2})(\frac{L}{2}) = 0$. After solving for Wx and factoring out $\frac{L}{2}$ out of the right side of the equation, I obtain the equation as follows: $Wx = \frac{L}{2}(\frac{2qQ}{4\pi\epsilon_0 h^2} + W - \frac{qQ}{4\pi\epsilon_0 h^2})$. Now after combining like terms and dividing each side by the weight variable $W$, we get the final solution for x which is as follows: $x= (\frac{L}{2})(\frac{qQ}{4\pi\epsilon_0 h^2W} + 1)$