Unit 5: Magnetic Fields and Electromagnetism

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Magnetism describes forces on moving charges and currents. Unlike electric fields, magnetic fields do no work on a point charge because the magnetic force is always perpendicular to the charge’s velocity. This unit focuses on Lorentz forces, fields produced by currents, and Ampere’s law.

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Useful variables

• $\vec{B}$ = magnetic field (Units: tesla, $\text{T}$)
• $q$ = charge
• $\vec{v}$ = velocity
• $I$ = current
• $L$ = length of wire in a magnetic field
• $\mu_0$ = permeability of free space = $4\pi\times10^{-7}\ \text{T}\cdot\text{m/A}$
• $\Phi_B$ = magnetic flux

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Magnetic Force on a Moving Charge

A charge moving through a magnetic field feels the magnetic part of the Lorentz force:

$$\vec{F}_B = q\vec{v}\times\vec{B}.$$

The magnitude is

$$F_B = \lvert q \rvert vB\sin\theta.$$

The force is perpendicular to both $\vec{v}$ and $\vec{B}$. For positive charges, use the right-hand rule for $\vec{v}\times\vec{B}$. For negative charges, reverse the direction.

The full electric and magnetic force is

$$\vec{F}=q\vec{E}+q\vec{v}\times\vec{B}.$$

Since $\vec{F}_B$ is always perpendicular to $\vec{v}$, it does no work: $P = \vec{F}_B\cdot\vec{v} = q(\vec{v}\times\vec{B})\cdot\vec{v} = 0$, since $\vec{v}\times\vec{B}$ is perpendicular to $\vec{v}$. A magnetic field can change a charge’s direction but never its speed.

Example. A proton ($q = +1.6\times10^{-19}\ \text{C}$) moves at $v = 4.0\times10^5\ \text{m/s}$ in the $+x$ direction through a uniform field $\vec{B} = 0.50\ \text{T}$ pointing in the $+y$ direction. Find the magnitude and direction of the magnetic force.

The velocity is perpendicular to the field ($\theta = 90^\circ$), so the magnitude is

$$F_B = \lvert q\rvert vB\sin 90^\circ = (1.6\times10^{-19})(4.0\times10^5)(0.50) = 3.2\times10^{-14}\ \text{N}.$$

For the direction, evaluate $\vec{v}\times\vec{B} = (v\hat{x})\times(B\hat{y}) = vB\,(\hat{x}\times\hat{y}) = vB\,\hat{z}$. Since the charge is positive, $\vec{F}_B$ points in $+z$ (out of the page if $x$ is right and $y$ is up). For an electron with the same velocity, the force would point in $-z$.

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Charged Particles in Uniform Magnetic Fields

If a charge moves perpendicular to a uniform magnetic field, the magnetic force provides centripetal force:

$$\lvert q \rvert vB=\frac{mv^2}{r}.$$

Thus

$$r=\frac{mv}{\lvert q\rvert B}.$$

The angular frequency is

$$\omega=\frac{\lvert q \rvert B}{m},$$

and the period is

$$T=\frac{2\pi m}{\lvert q \rvert B}.$$

If the velocity has a component parallel to $\vec{B}$, that component is unchanged and the path becomes a helix.

Notice that the period $T = 2\pi m/(\lvert q\rvert B)$ does not depend on the speed or the radius: a faster particle simply travels a proportionally larger circle in the same time. This speed-independence is the principle behind the cyclotron.

Example. An electron ($m = 9.11\times10^{-31}\ \text{kg}$, $\lvert q\rvert = 1.6\times10^{-19}\ \text{C}$) moves at $v = 2.0\times10^6\ \text{m/s}$ perpendicular to a uniform field $B = 1.5\times10^{-3}\ \text{T}$. Find the radius of its circular path, the period, and the frequency.

The magnetic force supplies the centripetal force, $\lvert q\rvert vB = mv^2/r$, so

$$r = \frac{mv}{\lvert q\rvert B} = \frac{(9.11\times10^{-31})(2.0\times10^6)}{(1.6\times10^{-19})(1.5\times10^{-3})} = \frac{1.82\times10^{-24}}{2.4\times10^{-22}} \approx 7.6\times10^{-3}\ \text{m}.$$

The period is independent of $v$:

$$T = \frac{2\pi m}{\lvert q\rvert B} = \frac{2\pi(9.11\times10^{-31})}{(1.6\times10^{-19})(1.5\times10^{-3})} = \frac{5.72\times10^{-30}}{2.4\times10^{-22}} \approx 2.4\times10^{-8}\ \text{s}.$$

The frequency (the cyclotron frequency) is

$$f = \frac{1}{T} = \frac{\lvert q\rvert B}{2\pi m} \approx 4.2\times10^{7}\ \text{Hz}.$$

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Velocity Selectors and Mass Spectrometers

When electric and magnetic forces oppose each other,

$$qE=qvB.$$

Only particles with speed

$$v=\frac{E}{B}$$

pass through undeflected. This is the basic idea of a velocity selector.

After velocity selection, a magnetic field can separate particles by mass-to-charge ratio because

$$r=\frac{mv}{\lvert q \rvert B}.$$

Example. A velocity selector uses crossed fields $E = 3.0\times10^4\ \text{V/m}$ and $B_1 = 0.10\ \text{T}$. The selected beam (singly charged ions, $q = 1.6\times10^{-19}\ \text{C}$) then enters a mass-spectrometer region with field $B_2 = 0.20\ \text{T}$. Find the selected speed, and the radii of two isotopes of mass $m_1 = 3.32\times10^{-26}\ \text{kg}$ and $m_2 = 3.65\times10^{-26}\ \text{kg}$.

In the selector, the electric and magnetic forces balance, $qE = qvB_1$, so only ions with

$$v = \frac{E}{B_1} = \frac{3.0\times10^4}{0.10} = 3.0\times10^5\ \text{m/s}$$

pass straight through. In the spectrometer the path radius is $r = mv/(qB_2)$:

$$r_1 = \frac{(3.32\times10^{-26})(3.0\times10^5)}{(1.6\times10^{-19})(0.20)} = \frac{9.96\times10^{-21}}{3.2\times10^{-20}} \approx 0.311\ \text{m},$$ $$r_2 = \frac{(3.65\times10^{-26})(3.0\times10^5)}{(1.6\times10^{-19})(0.20)} = \frac{1.095\times10^{-20}}{3.2\times10^{-20}} \approx 0.342\ \text{m}.$$

The heavier isotope curves on the larger circle. The two beams land $2(r_2 - r_1) \approx 6.2\ \text{cm}$ apart on the detector, which is how a mass spectrometer resolves isotopes.

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Magnetic Force on a Current-Carrying Wire

A wire segment carrying current $I$ in a magnetic field feels

$$\vec{F}=I\vec{L}\times\vec{B},$$

where $\vec{L}$ points in the direction of conventional current and has magnitude equal to the length of wire in the field.

The magnitude is

$$F=ILB\sin\theta.$$

This force comes from the magnetic forces on the moving charges inside the wire: each carrier feels $q\vec{v}_d\times\vec{B}$, and summing over all carriers in a length $L$ (with $I = nqv_d A_{\text{cross}}$) gives $\vec{F} = I\vec{L}\times\vec{B}$.

Example. A straight wire of length $L = 0.40\ \text{m}$ carries current $I = 8.0\ \text{A}$ in the $+x$ direction. It sits in a uniform field $\vec{B} = 0.25\ \text{T}$ pointing into the page ($-z$, with $x$ to the right and $y$ up). Find the magnitude and direction of the force.

The current direction ($+x$) is perpendicular to $\vec{B}$ ($-z$), so $\theta = 90^\circ$ and

$$F = ILB\sin 90^\circ = (8.0)(0.40)(0.25) = 0.80\ \text{N}.$$

For the direction, $\vec{L}\times\vec{B} = (L\hat{x})\times(-B\hat{z}) = -LB\,(\hat{x}\times\hat{z}) = -LB(-\hat{y}) = +LB\,\hat{y}$. So the force points in $+y$ (upward). Check with the right-hand rule: fingers point along the current ($+x$), curl toward $\vec{B}$ (into page), thumb points up. The force is $0.80\ \text{N}$ upward.

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Torque on a Current Loop

A current loop in a magnetic field experiences a torque. For a loop with $N$ turns, current $I$, area $A$, and magnetic moment

$$\vec{\mu}=NI\vec{A},$$

the torque is

$$\vec{\tau}=\vec{\mu}\times\vec{B}.$$

Its magnitude is

$$\tau=NIAB\sin\theta.$$

The potential energy of a magnetic dipole in a uniform magnetic field is

$$U=-\vec{\mu}\cdot\vec{B} = -\mu B\cos\theta.$$

Here $\theta$ is the angle between $\vec{\mu}$ (the loop’s normal, by the right-hand rule) and $\vec{B}$. The energy is lowest when $\vec{\mu}$ is aligned with $\vec{B}$ ($\theta = 0$, $U = -\mu B$, stable) and highest when anti-aligned ($\theta = 180^\circ$, $U = +\mu B$, unstable). The torque always tries to rotate $\vec{\mu}$ into alignment with $\vec{B}$.

Example. A rectangular coil with $N = 50$ turns, sides $0.080\ \text{m}\times0.050\ \text{m}$, carries $I = 2.0\ \text{A}$ in a uniform field $B = 0.30\ \text{T}$. (a) Find the magnetic moment and the maximum torque. (b) Find the dipole energy when $\vec{\mu}$ is aligned with $\vec{B}$ and when it is perpendicular to $\vec{B}$.

The loop area is $A = (0.080)(0.050) = 4.0\times10^{-3}\ \text{m}^2$, so the magnetic moment is

$$\mu = NIA = (50)(2.0)(4.0\times10^{-3}) = 0.40\ \text{A}\cdot\text{m}^2.$$

The torque $\tau = NIAB\sin\theta = \mu B\sin\theta$ is maximum at $\theta = 90^\circ$ (plane of the loop parallel to $\vec{B}$):

$$\tau_{\max} = \mu B = (0.40)(0.30) = 0.12\ \text{N}\cdot\text{m}.$$

For the energy $U = -\mu B\cos\theta$: when $\vec{\mu}\parallel\vec{B}$ ($\theta = 0$),

$$U_{\parallel} = -\mu B = -(0.40)(0.30) = -0.12\ \text{J},$$

and when $\vec{\mu}\perp\vec{B}$ ($\theta = 90^\circ$),

$$U_{\perp} = -\mu B\cos 90^\circ = 0\ \text{J}.$$

Rotating the dipole from aligned to perpendicular requires an external agent to supply $\Delta U = 0 - (-0.12) = +0.12\ \text{J}$.

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Magnetic Fields from Currents

The magnetic field around a long straight wire is

$$B=\frac{\mu_0 I}{2\pi r}.$$

The direction follows the right-hand rule: thumb in the direction of conventional current, curled fingers show the magnetic field direction.

At the center of a circular loop of radius $R$,

$$B=\frac{\mu_0 I}{2R}.$$

For $N$ identical turns,

$$B=\frac{\mu_0 NI}{2R}.$$

Inside a long ideal solenoid,

$$B=\mu_0 nI,$$

where $n=N/L$ is turns per unit length.

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Biot-Savart Law

For a current element, the magnetic field contribution is

$$d\vec{B}=\frac{\mu_0}{4\pi}\frac{I\,d\vec{\ell}\times\hat{r}}{r^2}.$$

Use Biot-Savart when symmetry is not enough for Ampere’s law but the current geometry is integrable. The cross product means only the component of $d\vec{\ell}$ perpendicular to the line from source to field point contributes.

Proof (Field of an infinite straight wire from Biot–Savart). Place the wire along the $x$-axis carrying current $I$, and find the field at a point $P$ a perpendicular distance $r$ from the wire. Let a current element $I\,d\vec{\ell} = I\,dx\,\hat{x}$ sit at position $x$; the vector from the element to $P$ has magnitude $\sqrt{x^2 + r^2}$.

The magnitude of $d\vec{\ell}\times\hat{r}$ is $dx\sin\phi$, where $\phi$ is the angle between the wire and the line to $P$. From the geometry, $\sin\phi = r/\sqrt{x^2 + r^2}$. Every element contributes field in the same direction (out of the page, by the right-hand rule), so the magnitudes add:

$$dB = \frac{\mu_0 I}{4\pi}\frac{dx\sin\phi}{x^2 + r^2} = \frac{\mu_0 I}{4\pi}\frac{r\,dx}{(x^2 + r^2)^{3/2}}.$$

Integrate over the whole wire, $x$ from $-\infty$ to $+\infty$:

$$B = \frac{\mu_0 I r}{4\pi}\int_{-\infty}^{\infty}\frac{dx}{(x^2 + r^2)^{3/2}}.$$

The standard integral is

$$\int\frac{dx}{(x^2 + r^2)^{3/2}} = \frac{x}{r^2\sqrt{x^2 + r^2}},$$

which you can verify by the substitution $x = r\tan\phi$, $dx = r\sec^2\phi\,d\phi$, giving $\int \frac{r\sec^2\phi\,d\phi}{r^3\sec^3\phi} = \frac{1}{r^2}\int\cos\phi\,d\phi = \frac{\sin\phi}{r^2}$. Evaluating from $-\infty$ to $+\infty$, the factor $\dfrac{x}{\sqrt{x^2+r^2}}$ runs from $-1$ to $+1$:

$$\int_{-\infty}^{\infty}\frac{dx}{(x^2 + r^2)^{3/2}} = \frac{1}{r^2}\big[1 - (-1)\big] = \frac{2}{r^2}.$$

Therefore

$$B = \frac{\mu_0 I r}{4\pi}\cdot\frac{2}{r^2} = \frac{\mu_0 I}{2\pi r}.$$

Proof (On-axis field of a circular loop from Biot–Savart). A circular loop of radius $R$ carries current $I$. Find $\vec{B}$ at a point $P$ on the axis, a distance $x$ from the center. Each element $I\,d\vec{\ell}$ is perpendicular to the line of length $s = \sqrt{x^2 + R^2}$ joining it to $P$, so $\lvert d\vec{\ell}\times\hat{r} \rvert = d\ell$ and

$$dB = \frac{\mu_0 I}{4\pi}\frac{d\ell}{x^2 + R^2}.$$

By symmetry, the components of $d\vec{B}$ perpendicular to the axis cancel when summed around the loop; only the axial components survive. The axial component is $dB_x = dB\cos\alpha$, where $\cos\alpha = R/\sqrt{x^2 + R^2}$ (the angle between $d\vec{B}$ and the axis). Thus

$$dB_x = \frac{\mu_0 I}{4\pi}\frac{d\ell}{x^2 + R^2}\cdot\frac{R}{\sqrt{x^2 + R^2}} = \frac{\mu_0 I R}{4\pi (x^2 + R^2)^{3/2}}\,d\ell.$$

Everything in front of $d\ell$ is constant around the loop, so the integral just gives the circumference $\oint d\ell = 2\pi R$:

$$B = \frac{\mu_0 I R}{4\pi (x^2 + R^2)^{3/2}}(2\pi R) = \frac{\mu_0 I R^2}{2(x^2 + R^2)^{3/2}}.$$

At the center, $x = 0$:

$$B = \frac{\mu_0 I R^2}{2(R^2)^{3/2}} = \frac{\mu_0 I R^2}{2R^3} = \frac{\mu_0 I}{2R},$$

recovering the center-of-loop result.

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Ampere’s Law

Ampere’s law relates magnetic field circulation to enclosed current:

$$\oint \vec{B}\cdot d\vec{\ell}=\mu_0 I_{\text{enc}}.$$

It is most useful with high symmetry, such as long straight wires, solenoids, and toroids. Choose an Amperian loop so $\vec{B}$ is either constant and parallel to $d\vec{\ell}$ or perpendicular to it.

For a toroid with $N$ turns,

$$B=\frac{\mu_0 NI}{2\pi r}$$

inside the core, under the ideal toroid approximation.

Proof (Straight-wire field as an Ampère’s-law check). For an infinite straight wire, symmetry tells us $\vec{B}$ has constant magnitude on any circle of radius $r$ centered on the wire and points tangent to that circle (circular field lines). Choose such a circle as the Amperian loop. Then $\vec{B}\parallel d\vec{\ell}$ everywhere, and $B$ is constant, so

$$\oint\vec{B}\cdot d\vec{\ell} = B\oint d\ell = B(2\pi r).$$

The loop encloses the full current $I$, so Ampère’s law gives

$$B(2\pi r) = \mu_0 I \;\Rightarrow\; B = \frac{\mu_0 I}{2\pi r},$$

matching the Biot–Savart result with far less work — symmetry did the integral for us.

Proof (Solenoid field from Ampère’s law). An ideal solenoid has $n$ turns per unit length carrying current $I$, with a uniform axial field inside and (ideally) zero field outside. Choose a rectangular Amperian loop with one side of length $\ell$ inside the solenoid parallel to the axis, the opposite side outside, and two short sides crossing the wall.

Split $\oint\vec{B}\cdot d\vec{\ell}$ into the four legs:

• Inside leg (parallel to $\vec{B}$): contributes $B\ell$.
• Outside leg: $\vec{B} = 0$, contributes $0$.
• Two crossing legs: $\vec{B}$ is either zero (outside) or perpendicular to $d\vec{\ell}$ (inside), so each contributes $0$.

Therefore

$$\oint\vec{B}\cdot d\vec{\ell} = B\ell.$$

The current enclosed is the number of turns threading the loop, $n\ell$, each carrying $I$:

$$I_{\text{enc}} = n\ell I.$$

Ampère’s law then gives

$$B\ell = \mu_0 n\ell I \;\Rightarrow\; B = \mu_0 n I,$$

independent of position inside — confirming the field is uniform.

Example. A long cylindrical wire of radius $a$ carries a total current $I$ spread uniformly over its cross-section, so the current density is $J = I/(\pi a^2)$. Find $B$ at radius $r < a$ inside the conductor.

By symmetry the field is again circular, so for an Amperian circle of radius $r < a$, $\oint\vec{B}\cdot d\vec{\ell} = B(2\pi r)$. But this loop encloses only the fraction of current inside radius $r$:

$$I_{\text{enc}} = J\,(\pi r^2) = \frac{I}{\pi a^2}\,\pi r^2 = I\frac{r^2}{a^2}.$$

Ampère’s law gives

$$B(2\pi r) = \mu_0 I\frac{r^2}{a^2} \;\Rightarrow\; B = \frac{\mu_0 I r}{2\pi a^2}.$$

So inside the conductor $B\propto r$ (rising linearly from zero on the axis), and at the surface $r = a$ this matches $B = \mu_0 I/(2\pi a)$, the external field, as it must.

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Magnetic Forces Between Parallel Wires

Two long parallel wires carrying currents $I_1$ and $I_2$ separated by distance $d$ exert forces on each other. The force per unit length is

$$\frac{F}{L}=\frac{\mu_0 I_1I_2}{2\pi d}.$$

Currents in the same direction attract. Currents in opposite directions repel.

Proof (Force per length between parallel wires). Wire 1 carries $I_1$ and produces, at the location of wire 2 (distance $d$ away), a field of magnitude

$$B_1 = \frac{\mu_0 I_1}{2\pi d}.$$

This field is perpendicular to wire 2. The force on a length $L$ of wire 2 (carrying $I_2$) is $\vec{F} = I_2\vec{L}\times\vec{B}_1$, with magnitude (since $\vec{L}\perp\vec{B}_1$, $\sin\theta = 1$):

$$F = I_2 L B_1 = I_2 L\frac{\mu_0 I_1}{2\pi d}.$$

Dividing by $L$,

$$\frac{F}{L} = \frac{\mu_0 I_1 I_2}{2\pi d}.$$

The result is symmetric in $I_1$ and $I_2$, as Newton’s third law requires. For the direction, apply the right-hand rule: if the currents are parallel (same direction), $\vec{L}\times\vec{B}_1$ points from wire 2 toward wire 1, so they attract; antiparallel currents repel.

Example. Two parallel wires $d = 0.020\ \text{m}$ apart carry $I_1 = 10\ \text{A}$ and $I_2 = 15\ \text{A}$ in the same direction. Find the force per unit length and state whether it is attractive or repulsive.

$$\frac{F}{L} = \frac{\mu_0 I_1 I_2}{2\pi d} = \frac{(4\pi\times10^{-7})(10)(15)}{2\pi(0.020)} = \frac{(2\times10^{-7})(10)(15)}{0.020} = \frac{3.0\times10^{-5}}{0.020} = 1.5\times10^{-3}\ \text{N/m}.$$

Here I used $\dfrac{\mu_0}{2\pi} = \dfrac{4\pi\times10^{-7}}{2\pi} = 2\times10^{-7}\ \text{T}\cdot\text{m/A}$. Since the currents run the same direction, the force is attractive: $1.5\times10^{-3}\ \text{N}$ per meter pulling the wires together.

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Magnetic Flux

Magnetic flux through a surface is

$$\Phi_B=\int \vec{B}\cdot d\vec{A}.$$

For a uniform field through a flat surface,

$$\Phi_B=BA\cos\theta.$$

Magnetic flux becomes central in electromagnetic induction.

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Key equations

Idea	Equation
Lorentz force	$\vec{F} = q\vec{E} + q\vec{v}\times\vec{B}$
Magnetic force magnitude	$F = \lvert q\rvert vB\sin\theta$
Radius in a $B$ field	$r = \dfrac{mv}{\lvert q\rvert B}$
Period / cyclotron frequency	$T = \dfrac{2\pi m}{\lvert q\rvert B}, \quad f = \dfrac{\lvert q\rvert B}{2\pi m}$
Velocity selector	$v = \dfrac{E}{B}$
Force on a wire	$\vec{F} = I\vec{L}\times\vec{B}, \quad F = ILB\sin\theta$
Magnetic moment	$\vec{\mu} = NI\vec{A}$
Torque on a loop	$\vec{\tau} = \vec{\mu}\times\vec{B}, \quad \tau = NIAB\sin\theta$
Dipole energy	$U = -\vec{\mu}\cdot\vec{B} = -\mu B\cos\theta$
Field of straight wire	$B = \dfrac{\mu_0 I}{2\pi r}$
Field on axis of loop	$B = \dfrac{\mu_0 I R^2}{2(x^2 + R^2)^{3/2}}$
Field at center of loop	$B = \dfrac{\mu_0 NI}{2R}$
Field inside solenoid	$B = \mu_0 nI$
Field inside toroid	$B = \dfrac{\mu_0 NI}{2\pi r}$
Biot–Savart law	$d\vec{B} = \dfrac{\mu_0}{4\pi}\dfrac{I\,d\vec{\ell}\times\hat{r}}{r^2}$
Ampère’s law	$\oint\vec{B}\cdot d\vec{\ell} = \mu_0 I_{\text{enc}}$
Force between parallel wires	$\dfrac{F}{L} = \dfrac{\mu_0 I_1 I_2}{2\pi d}$
Magnetic flux	$\Phi_B = \int\vec{B}\cdot d\vec{A} = BA\cos\theta$

Practice

FRQ

1. Temporary placeholder FRQ for wiring/testing — replace with a real free-response question for this unit.

   $(A)$ State one key idea from this unit and explain it in your own words.

   $(B)$ Give a worked example or application of that idea.