Steady Magnetic Field¶

Biot-Savart Law¶

The Ampere's Law is \(\(d\vec{F}_{12} = \frac{\mu_0}{4\pi} \frac{i_2ds_2 \times (i_1ds_1 \times \hat{r}_{12})}{r_{12}^2}\)\)

with \(\mu_0 = 4\pi\times 10^{-7} N/A^2\). \(F_{12}\) and \(F_{21}\) may not be equal.

The Magnetic Induction Strength is \(\(\vec{B} = \frac{\mu_0}{4\pi} \oint_{L} \frac{id\vec{s} \times \hat{r}}{r^2}\)\)

For a section of line,

\[ \begin{gathered} x = -r_0\cot\theta \Rightarrow dx = \frac{r_0d\theta}{\sin^2\theta} \\\\ B = \int_{\theta_1}^{\theta_2} \frac{\mu_0i}{4\pi} \frac{\sin\theta \cdot \dfrac{r_0d\theta}{\sin^2\theta}}{(r_0/\sin\theta)^2} = \frac{\mu_0i}{4\pi r_0} \int_{\theta_1}^{\theta_2} \sin\theta d\theta = \frac{\mu_0i}{4\pi r_0} (\cos\theta_1 - \cos\theta_2) \end{gathered} \]

When the line is infinite, \(B = \dfrac{\mu_0i}{2\pi r_0}\).

For a circular loop, \(\(B = \frac{\mu_0i}{4\pi} \oint \frac{ds}{(z/\sin\theta)^2} \cos\theta = \frac{\mu_0iR}{2z^2} \sin^2\theta \cos\theta = \frac{\mu_0}{2} \frac{iR^2}{(R^2 + z^2)^{3/2}}\)\)

When at center, \(B = \dfrac{\mu_0i}{2R}\).

When \(z >> R\), \(B = \dfrac{\mu_0iR^2}{2z^3} = \dfrac{\mu_0i\pi R^2}{2\pi z^3} = \dfrac{\mu_0iA}{2\pi z^3}\).

Then we can define magnetic dipole moment as

\[ \begin{gathered} \mu = iA = i\pi R^2 \\\\ B = \frac{\mu_0}{2\pi} \frac{\mu}{r^3} \end{gathered} \]

For a flat strip of copper with negligible thickness

\[ \begin{gathered} d = \frac{R}{\cos\theta} \\\\ dB_x = dB \cdot \cos\theta = \frac{\mu_0 \dfrac{i}{a} dx}{2\pi d} \cdot \cos\theta \\\\ x = R\tan\theta \Rightarrow dx = \frac{Rd\theta}{\cos^2\theta} \\\\ \begin{aligned} B_x &= \frac{\mu_0 i}{2\pi aR} \int \cos^2\theta dx = \frac{\mu_0 i}{2\pi a} \int_{-\alpha}^{\alpha} d\theta \\ &= \frac{\mu_0 i}{\pi a} \alpha = \frac{\mu_0 i}{\pi a} \tan^{-1} \frac{a}{2R} \end{aligned} \end{gathered} \]

When \(R >> a\), \(B = \dfrac{\mu_0 i}{2\pi R}\).

When \(R \rightarrow 0\), \(B = \dfrac{\mu_0 i}{2a}\).

Let the length of solenoid be \(L\), the radius as \(R\), the number of turns per unit length as \(n\), then

\[ \begin{gathered} B = \frac{\mu_0}{2} \int_{- \frac{L}{2}}^{\frac{L}{2}} \frac{iR^2 \cdot ndl}{[R^2 + (x - l)^2]^{3/2}} \\\\ \frac{x - l}{R} = \cot\beta \Rightarrow dl = \frac{R}{\sin^2\beta} d\beta \\\\ \begin{aligned} B &= \frac{\mu_0}{2} \int_{-\beta_1}^{\beta_2} \frac{R^2 ni \dfrac{R}{\sin^2\beta} d\beta}{\left( \dfrac{R^2}{\sin^2\beta} \right)^{3/2}} \\ &= \frac{\mu_0}{2} \cdot ni \int_{-\beta_1}^{\beta_2} \sin\beta d\beta \\ &= \frac{1}{2} \mu_0ni (\cos\beta_1 - \cos\beta_2) \end{aligned} \end{gathered} \]

When \(L\rightarrow \infty\), \(\beta_1 = 0, \beta = \pi, B = \mu_0ni\).

At the end of solenoid, \(\beta_1 = 0, \beta_2 = \dfrac{\pi}{2}, B = \frac{1}{2} \mu_0ni\).

For a solenoid with many layers,

\[ \begin{aligned} B &= \int_{R_1}^{R_2} \frac{1}{2} \mu_0 \frac{Ni}{2l(R_2 - R_1)} \cdot \frac{2l}{\sqrt{l^2 + r^2}} dr \\ &= \mu_0 jl \int_{R_1}^{R_2} \frac{dr}{\sqrt{l^2 + r^2}} \\ &= \mu_0 jl \ln \frac{R_2 + \sqrt{R_2^2 + l^2}}{R_1 + \sqrt{R_1^2 + l^2}} \\ \end{aligned} \]

The Gauss’Law and Ampere’s Loop Law of a magnetic field¶

The Gauss' Law of s magnetic field is \(\(\oiint B\bullet dA = 0\)\)

The Ampere’s Loop Law of a magnetic field is \(\(\oint B\bullet dl = \mu_0 \sum_{in\ loop} i\)\)

So the magnetic field inside a long wire is \(\(B = \mu_0 i \cdot \frac{\pi r^2}{\pi R^2} \cdot \frac{1}{2\pi r} = \frac{\mu_0 ir}{2\pi R^2}\)\)

Consider an infinite current sheet, suppose \(n\) is the number of wires per length, then \(\(B = \frac{\mu_0 \cdot wni}{2w} = \frac{1}{2} \mu_0 ni\)\)

Therefore, when inside a solenoid, \(B = \mu_0 ni\), when outside a solenoid, \(B = 0\).

For a toroid, \(\(B = \frac{\mu_0 Ni}{2\pi r} = \mu_0 ni\)\)

The magnetic force on a carrying-current wire¶

The Ampere’s Force is \(\(dF = ids \times B\)\)

In a uniform magnetic field, if the starting point and end point of a wire is determined, then the Ampere force generated by different paths is equal.

For a rectangular loop of wire

\[ \begin{aligned} \tau &= F_{AB} \cdot \frac{b}{2} \sin\theta + F_{CD} \cdot \frac{b}{2} \sin\theta \\ &= iaB \cdot \frac{b}{2} \sin\theta + iaB \cdot \frac{b}{2} \sin\theta \\ &= iabB\sin\theta \\ &= iBA\sin\theta \end{aligned} \]

For arbitrary shape loop, split it into thin ladders, we can similarly get \(\tau = iBA\).

Let's define \(\vec{\mu} = iA\vec{n}\), then \(\vec{\tau} = \vec{\mu} \times \vec{B}\).

If we define the potential energy of a magnetic dipole is 0 when \(\theta = \pi/2\), then \(U = - \vec{\mu} \bullet \vec{B}\)

The motion of a charge in a magnetic field¶

The Lorentz Force is \(\(F = qv \times B\)\)

Mass Spectrometer¶

\[ qvB = m \frac{v^2}{R} \\\\ \Rightarrow R = \frac{mv}{qB} \]

\[ \frac{1}{2}mv^2 = qV \\\\ \Rightarrow \frac{m}{q} = \frac{R^2B^2}{2V} \]

Cyclotron¶

\[T = \frac{2\pi R}{v} = \frac{2\pi m}{B}\]

\(T\) is independent of \(v\).

\[ E_K = \frac{1}{2}m \left( \frac{BRq}{m} \right)^2 = \frac{B^2R^2q^2}{2m} \]

Hall Effect¶

\[ j = \frac{dq}{dt\cdot A} = \frac{qnA\cdot dl}{dt\cdot A} = qn\cdot \frac{dl}{dt} = qnv \]

\[ \begin{aligned} V &= E\cdot b = vB \cdot b = \frac{j}{nq} \cdot B \cdot b \\ &= \frac{(j\cdot b\cdot d)\cdot B}{nq\cdot d} = \frac{iB}{nq\cdot d} \\ &= \frac{1}{nq} \cdot \frac{iB}{d} \end{aligned} \]

So \(V = \kappa \cdot \dfrac{iB}{d}\), where \(d\) is the thickness along magnetic field.

Shorthand¶

Biot-Savart Law
Magnetic field around an infinite wire
Magnetic field around a flat strip of copper
Magnetic field along the axis of a circular loop
Magnetic field along the axis of a solenoid
Magnetic field around an infinite current sheet (Ampere's Loop Law)
Torque of a loop in a uniform magnetic field
Energy of a loop in a uniform magnetic field
Lorentz Force (mind the charge polarity)
Hall Effect