Magnetic field in the hydrogen atom with Bohr theory

Question:
Use Bohr theory. Find the magnetic field by electron's exerting at the nucleus of the hydrogen atom. The radius of hydrogen is assumed to be 0.529 $\times$ 10$^{-10}$ m.

Answer:
We can assume that the electron moves around the nucleus as making a circular current. According to the Biot-Savart Law, we have the magnetic field of a circular current as
\[
B_z = \frac{\mu_0 I}{4 \pi}\int^{2\pi r}_{0}\frac{dl}{r^2}=\frac{\mu_0 I}{2r}
\]
Now, we need to obtain the current. The electric current is defined by $I=\frac{Q}{T}$. The charge of an electron or a proton is 1.61 $\times$ 10$^{-19}$ C. In order to get the current, we have to find the time period. That is
\[
T=\frac{2\pi r}{v}
\]
where $v$ is the tangential velocity of the electron. So the current is expressed by
\[
I = \frac{Q}{\frac{2\pi r}{v}}
\]
In terms of the classical theory, the centripetal force is equal to the Coulomb force. Then, solve for the velocity.
\begin{eqnarray}
\frac{mv^2}{r}&=&\frac{ke^2}{r^2} \\
v&=&\sqrt{\frac{ke^2}{mr}}  \\
  &=& \sqrt{\frac{8.99 \times 10^9 (1.61 \times 10^{-19})^2}{9.11 \times 10^{-31}\cdot 0.529 \times 10^{-10}}}  \\
  &=& 2.20 \times 10^6 \ \mathrm{m/s}
\end{eqnarray}
Therefore, we calculate the current.
\[
I=\frac{Q}{\frac{2\pi r}{v}}=\frac{1.61 \times 10^{-19}}{\frac{2\pi \times 0.529\times 10^{-10}}{2.20\times 10^6}}=1.07\times 10^{-3} \ \mathrm{A}
\]
Then, we can find the magnetic field.
\[
B=\frac{\mu_0 I}{2r}=\frac{4\pi \times 10^{-7} \times1.07\times 10^{-3}}{2 \times 0.529\times 10^{-10}}= 12.7 \ \mathrm{T}
\]

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Ampere's law: A toroid carrying current

Question:
A toroid has $N$ turns both sides carrying equal current $I$. The inner and outer radii are $a$ and $b$, respectively. A distance $r$ is supposed to be exact the middle between $a$ and $b$. Find the magnetic field at $r$.

Answer:
From Ampere's law, if inside the contour has a current $I$, we have the relationship:
\[
\oint \vec{B}\cdot dl = \mu_0 I
\]
The radius of the contour is $r$ which encloses the current $N\times I$. Therefore,
\[
B(2\pi r) = \mu_0 NI
\]
Recall that $r=\frac{a+b}{2}$; thus, the magnetic field is
\[
B = \frac{\mu_0 NI}{\pi(a+b)}
\]

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Magnetic fields from current flows and resistor circuit

Question:
The wire splits into two ways, which are bent circularly with radius of $a$. The upper half wire has resistance $2R$ and the lower part has $R$. Find the magnetic fields at the center of the circle in terms of the total current $I$.

Answer:
The magnetic field created by current $I_1$ is denoted as $B_1$. From Biot-Savart law,
\begin{eqnarray}
\vec{B_1} &=& \oint \frac{\mu_0}{4\pi}\frac{I d\vec{l}\times \vec{r}}{r^2}  \\
    &=& \frac{\mu_0}{4\pi}\frac{I_1}{a^2}\int_{0\rightarrow \pi a}dl (-\vec{z})  \\
    &=& -\frac{\mu_0}{4\pi}\frac{I_1}{a^2}\pi a\vec{z} \\
    &=& -\frac{\mu_0 I_1}{4a}\vec{z}
\end{eqnarray}
Likewise, we can obtain the magnetic field created by the lower part.
\begin{equation}
\vec{B_2} = \frac{\mu_0 I_2}{4a}\vec{z}
\end{equation}
This is a parallel connection, and we can use
\[
\frac{I_1}{I_2}=\frac{R}{2R} \quad \rightarrow \quad I_1=\frac{1}{2}I_2
\]
The current is conserved.
\[
I = I_1 + I_2 \quad \rightarrow \quad I_2 = I - I_1
\]
Therefore,
\[
I_1=\frac{1}{2}(I - I_1) \quad \rightarrow \quad I_1 = \frac{I}{3}
\]
Hence, the other current will be
\[
I_2 = \frac{2I}{3}
\]
The magnetic field is expressed as
\[
\vec{B}=\vec{B_1}+\vec{B_2}=\frac{\mu_0}{4a}(I_2-I_2)\vec{z}=\frac{\mu_0}{4a}\left(\frac{2I}{3}-\frac{I}{3}\right)\vec{z}=\frac{\mu_0 I}{12a}\vec{z}
\]
The magnetic field is directed toward us.

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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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