Magnetic field by a circular ring with current flow

Question:
As shown in the figure, there is a circular coil with current flow, $I$. Knowing that the radius is $r$, find the magnetic field at the center of the coil.

Answer:
Use the Biot-Savart Law.
\[
d\vec{B} = \frac{\mu_0 I}{4 \pi}\frac{d\vec{l}\times \vec{r}}{r^3}
\]
where $d\vec{l}$ is the infinitesimal length of the coil. The cross-product can be written as $d\vec{l}\times \vec{r} = |d\vec{l}|r\sin\theta$ The equation becomes
\[
d\vec{B} = \frac{\mu_0 I}{4 \pi}\frac{|d\vec{l}|\sin\theta}{r^2}
\]
$\theta$ is the angle between the radius and length vectors. They are always perpendicular, so $\theta = \frac{\pi}{2}$. Thus, $\sin\frac{\pi}{2}=1$. Now integrate it with the circular contour.
\[
\vec{B} = \oint_C \frac{\mu_0 I}{4 \pi}\frac{|d\vec{l}|}{r^2}
\]
All the constants can be placed outside the integral.
\[
\vec{B} = \frac{\mu_0 I}{4 \pi r^2} \oint_C |d\vec{l}|
\]
The contour integral gives $2\pi r$. Therefore,
\[
\vec{B} = \frac{\mu_0 I}{4 \pi r^2}2 \pi r = \frac{\mu_0 I}{2 r}
\]

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Factor theorem

Question:
If the polynomial, $P(x) = ax^4 + (b-a)x^3 + (1-2ab)x^2 + (ab-10)x + 2ab$, has a factor $x-2$, find $a$ and $b$.

Answer:
The factor theorem says, "If a polynomial, $P(x)$, has a factor, $x-\alpha$, then $P(\alpha) = 0$. In this problem, we can put $P(2) = 0$. Thus,
\begin{eqnarray*}
P(2) &=& 16a + 8(b-a) + 4(1-2ab) + (ab-10) + 2ab = 0   \\
&\rightarrow& 4ab - 8a - 8b + 16 = 0    \\
&\rightarrow& ab - 2a - 2b + 4 = 0    \\
&\rightarrow& a(b - 2) - 2(b - 2) = 0    \\
&\rightarrow& (a-2)(b-2) = 0    \\
\end{eqnarray*}
In order to have the factor $x-2$ for $P(x)$, $a=2$ and $b=2$.

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Conservation of angular momentum

Question:
A solid cylinder with a moment of inertia, $I_0$, rotates about its center at an angular velocity of $\omega_0$. Another solid cylinder, which is initially rest, is put onto the rotating one gently. Both eventually rotate at an angular velocity, $\omega_f$. Find the velocity $\omega_f$.

Answer:
The angular momentum is defined as
\[
L = I\omega
\]
With the constant velocity, initial and final angular momenta are conserved. Namely, initial total momentum = final total momentum. In this case, we have
\[
I_0 \omega_0 = (I_0 + I_1)\omega_1
\]
Solve for $\omega_f$.
\[
\omega_f = \frac{I_0 \omega_0}{I_0 + I_1}
\]

Torques and the balance

Question:
A horizontal beam is attached to a wall. The length and weight are 10.0 m and 200 N, respectively. The far end is supported by a wire and the angle between the beam and wire is 60$^o$. A person who has 500-N weight stands 2.00 m from the wall. What is the tension in the wire?

Answer:
This system is at equilibrium, so the sum of all of the torques must be zero.
\[
\sum \tau = 0
\]
The torque is defined as
\[
\tau = rF\sin \theta
\]
where $r$ and $F$ are the lever arm and the force. The angle $\theta$ is taken from the beam. The weight of the beam acts on the middle of the length by assuming that the mass is uniformly distributed. Plug in the numbers to calculate the net torque:
\begin{eqnarray*}
500\cdot 2.00 \sin(270^o) &+& 200\cdot (5.00)\sin(270^o) + T\cdot 10.0 \sin(60^o) = 0    \\
- 1000 - 1000 &+& 10.0T\sin(60^o)  = 0     \\
T &=& \frac{2000}{10\sin(60^o)}       \\
T &=& 231 \quad \mathrm{N}
\end{eqnarray*}

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Quadratic equation: The relationship between solutions and coefficients 2

Question:
The equation, $kx^2 -(k+3)x - 1 =0$, which has real coefficients, has complex roots, $a+ib$ and $a-ib$. In order to have those roots, find the range of $k$. If they are pure imaginary roots, find the value of $k$.

Answer:
The discriminant of the quadratic equation, $ax^2+bx+c=0$, is
\[
D = b^2 - 4ac
\]
For complex roots, we must have $D<0$ and $k \neq 0$. Thus,
\begin{eqnarray*}
D &=& (k+3)^2 +4k    \\
    &=& k^2+10k+9      \\
    &=& (k+1)(k+9)  < 0
\end{eqnarray*}
The range of $k$ is
\[
-9 < k < -1 \quad \mathrm{but} \quad k \neq 0
\]
If the roots are pure imaginary, we let the roots be $ib$ and $-ib$. We have the following relationship:
\[
ib + (-ib) = - \frac{-(k+3)}{k}
\]
This is from the relationship between the roots and coefficients. Then,
\begin{eqnarray*}
0 &=& k +3  \\
\rightarrow k &=& -3
\end{eqnarray*}
This satisfies $-9 < k < -1$.

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Quadratic equation: The relationship between solutions and coefficients 1

Question:
The roots of $x^2 - 5x + 3 = 0$ are $\alpha$ and $\beta$. Another quadratic equation $x^2 + px + q =0$ has roots, $\alpha^3$ and $\beta^3$. Find $p$ and $q$.

Answer:
If $\alpha$ and $\beta$ are the roots of a quadratic equation, $ax^2 + bx + c = 0 \quad (a\neq 0)$, the coefficients and roots have the following relationship:
\[
\alpha + \beta = -\frac{b}{a}, \\
\alpha \cdot \beta = \frac{c}{a}
\]
From the first equation, we have
\[
\alpha + \beta = 5, \\
\alpha \cdot \beta = 3
\]
From the other equation, they can be expressed as
\[
\alpha^3+\beta^3 = -p, \\
(\alpha\beta)^3 = q
\]
We can easily have
\[ q = 27 \]
However, $\alpha^3+\beta^3$ should be modified as follows:
\begin{eqnarray*}
\alpha^3+\beta^3 &=& (\alpha+\beta)(\alpha^2 -\alpha\beta + \beta^2)  \\
                             &=& (\alpha+\beta)(\alpha^2 +2\alpha\beta + \beta^2 - 3\alpha\beta)  \\
                            &=& (\alpha+\beta)((\alpha + \beta)^2 - 3\alpha\beta)
\end{eqnarray*}
Therefore, we obtain
\[
p = - (\alpha+\beta)((\alpha + \beta)^2 - 3\alpha\beta) = 5 \cdot (5^2 - 3 \cdot 3) = -80
\]

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The bounded solution of a differential equation

Question:
Find the bounded solution of the following ordinary differential equation:
\[ \frac{dx}{dt} = 2x + \sin t \]
for $-\infty < t < \infty$.

Answer:
In order to obtain the general solution, consider
\[ \frac{dx}{dt}-2x_g = 0 \]
Let us put $D \equiv \frac{d}{dt}$ as an operator. Thus, the above will become:
\[ (D - 2)x_g = 0 \]
Thus, we can have
\[ x_g = Ce^{2t} \]
Now, consider the particular solution. Use the operator to express the whole equation.
\[ (D-2)x_p = \sin t \]
$\sin t$ can in general be replaced with exponential function $e^{it}$, and the imaginary part of the final result will be taken.
\begin{eqnarray*}
x_p &=& \frac{1}{D-2}e^{it}   \\
   &=&  \frac{1}{i-2}e^{it}   \\
   &=&  \frac{-1}{2-i}e^{it}  \\
   &=&  \frac{-1}{2-i}\frac{2+i}{2+i}e^{it}  \\
   &=&  \frac{-(2+i)}{4+1}e^{it}  \\
   &=&  \frac{-(2+i)}{5}e^{it}  \\
   &=&  \frac{-2-i}{5}(\cos t + i\sin t)  \\
   &=&   \left(-\frac{2}{5}\cos t + \frac{1}{5}\sin t \right) + i \left(-\frac{1}{5}\cos t - \frac{2}{5}\sin t \right)
\end{eqnarray*}
The particular solution is the imaginary part.
\[ x_p = -\frac{1}{5}\cos t - \frac{2}{5}\sin t \]
The entire solution is
\[ x = Ce^{2t}-\frac{1}{5}\cos t - \frac{2}{5}\sin t \]
The first term makes the solution diverged when $t \rightarrow \pm\infty$. Therefore, the term has to be eliminated. Namely, $C = 0$. The solution bounded is
\[ x = -\frac{1}{5}\cos t - \frac{2}{5}\sin t \]

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