Magnetic field of straight wire, circular loop, solenoid, toroid

The two master laws (WHY they exist)

WHY this form? It is the magnetic analogue of Coulomb's $1/r^2$ law, but the cross product $d\vec{l}\times\hat r$ encodes the fact that magnetic field circulates around the current rather than pointing away from it.

WHEN to use which? Use Ampère's law when the symmetry lets you pull $B$ out of the integral (straight wire, solenoid, toroid). Use Biot–Savart when there's no helpful symmetry (point on the axis of a loop).

1. Straight Wire — Ampère's Law (HOW)

Derivation from scratch: $\oint \vec B\cdot d\vec l = B\oint dl = B(2\pi r)$ Why this step? $B$ is constant in magnitude on the circle and parallel to $d\vec l$ everywhere, so the dot product is just $B\,dl$ .

Set equal to $\mu_0 I_{\text{enc}} = \mu_0 I$ : $B(2\pi r) = \mu_0 I \;\Rightarrow\; \boxed{B = \frac{\mu_0 I}{2\pi r}}$

Finite wire (Biot–Savart result) — for a segment subtending angles $\theta_1,\theta_2$ at the point: $B = \frac{\mu_0 I}{4\pi r}(\sin\theta_1 + \sin\theta_2)$ For an infinite wire $\theta_1=\theta_2=90°$ , giving back $\dfrac{\mu_0 I}{2\pi r}$ . ✓

2. Circular Loop (on axis) — Biot–Savart (HOW)

No simple Amperian loop works, so integrate.

Derivation: Each element: $d\vec l \perp \hat r$ , so $|d\vec l \times \hat r| = dl$ . Distance $r=\sqrt{a^2+x^2}$ . $dB = \frac{\mu_0 I\,dl}{4\pi (a^2+x^2)}$ Axial part: multiply by $\cos\alpha = \dfrac{a}{\sqrt{a^2+x^2}}$ . $B = \int dB\cos\alpha = \frac{\mu_0 I}{4\pi}\frac{a}{(a^2+x^2)^{3/2}}\oint dl$ Why this step? $\oint dl = 2\pi a$ , the circumference. $\boxed{B = \frac{\mu_0 I a^2}{2(a^2+x^2)^{3/2}}}$

Figure — Magnetic field of straight wire, circular loop, solenoid, toroid

3. Solenoid — Ampère's Law (HOW)

Derivation: Take a rectangular Amperian loop of length $L$ , one side inside (field $B$ , parallel), opposite side far outside (field ≈ 0), two short sides perpendicular (contribute 0). $\oint \vec B\cdot d\vec l = BL$ Current enclosed = (turns in length $L$ ) × $I = nL\cdot I$ , where $n=N/L$ turns per metre. $BL = \mu_0 (nL) I \;\Rightarrow\; \boxed{B = \mu_0 n I}$

4. Toroid — Ampère's Law (HOW)

Derivation: Amperian loop = circle of radius $r$ inside the core. By symmetry $B$ constant on it: $\oint \vec B\cdot d\vec l = B(2\pi r)$ Enclosed current = $N I$ (loop crosses all $N$ turns once). $B(2\pi r) = \mu_0 N I \;\Rightarrow\; \boxed{B = \frac{\mu_0 N I}{2\pi r}}$

Common mistake Steel-manned common errors

❌ "Field inside a solenoid depends on its radius." Why it feels right: In a single loop the centre field $\mu_0 I/2a$ does depend on $a$ , so you expect radius to matter. Fix: In an ideal long solenoid, contributions from the whole length conspire to give $\mu_0 nI$ , independent of radius — the geometry is fundamentally different from one loop.

❌ "Toroid field is uniform like a solenoid." Why it feels right: Both come from $\mu_0 nI$ . Fix: Toroid has $B\propto 1/r$ , so it varies across the core thickness. Only thin cores look "uniform."

❌ Using Ampère's law for a circular loop's axial field. Why it feels right: Ampère's law is so easy elsewhere. Fix: No Amperian path has constant $B$ here → you must integrate (Biot–Savart).

❌ Forgetting $N$ for multi-turn coils. Always multiply single-turn results by the number of turns.

Recall Feynman: explain to a 12-year-old

Electricity flowing in a wire is like water rushing through a pipe, and it makes invisible "wind" that swirls around the pipe in circles. If you bend the pipe into a ring, the wind blows straight through the ring's hole. Stack lots of rings into a tube (a solenoid) and the wind inside becomes a strong, steady breeze — that's why it acts like a magnet. Bend that tube into a doughnut (a toroid) and the breeze gets trapped going round and round inside, with no wind escaping outside. How strong the breeze is just depends on how much current flows and how tightly the rings are packed.

Forecast-then-Verify checkpoint

Before reading answers, predict:

Double the current in a wire → field? (doubles)
Move twice as far from wire → field? (halves)
Pull a solenoid twice as long with same total turns → field inside? (halves, since $n$ halves)
Toroid field at inner vs outer edge → which bigger? (inner, since $B\propto1/r$ )

Flashcards

Biot–Savart law formula

d\vec B = \dfrac{\mu_0}{4\pi}\dfrac{I\,d\vec l\times\hat r}{r^2}

Ampère's law

\oint \vec B\cdot d\vec l = \mu_0 I_{\text{enc}}

Field of infinite straight wire at distance

r

B=\dfrac{\mu_0 I}{2\pi r}

Why use Ampère's law for a wire but Biot–Savart for a loop's axis?

Wire has circular symmetry letting

B

leave the integral; the loop axis has no Amperian path of constant

B

, so you must integrate.

Field at centre of circular loop radius

a

B=\dfrac{\mu_0 I}{2a}

Field on axis of loop, distance

x

B=\dfrac{\mu_0 I a^2}{2(a^2+x^2)^{3/2}}

Why do off-axis components cancel for a loop on its axis?

Diametrically opposite elements give perpendicular

d\vec B

components that are equal and opposite; only axial parts add.

Field inside a long solenoid

B=\mu_0 n I

, with

n=N/L

Does solenoid field depend on radius?

No — uniform and radius-independent for an ideal long solenoid.

Field at the end of a semi-infinite solenoid

\tfrac12\mu_0 n I

Field inside a toroid at radius

r

B=\dfrac{\mu_0 N I}{2\pi r}

Is the toroid field uniform?

No — it varies as

1/r

, stronger near the inner edge.

Field outside a toroid / in its hole

Zero (no enclosed current for those Amperian loops).

Right-hand grip rule

Thumb points along current, curled fingers give the magnetic field direction.

Finite straight wire field

B=\dfrac{\mu_0 I}{4\pi r}(\sin\theta_1+\sin\theta_2)

Value of

\mu_0

4\pi\times10^{-7}\ \text{T·m/A}

Connections

Biot-Savart Law — the integration engine behind loop & finite-wire fields.
Ampere's Law — symmetry shortcut for wire, solenoid, toroid.
Magnetic force on a current-carrying conductor — these fields exert forces.
Bar magnet and magnetic dipole — loop/solenoid behave as dipoles.
Faraday's law of induction — changing solenoid field induces EMF.
Magnetic flux — uniform solenoid field makes flux easy: $\Phi=BA$ .

Concept Map

Hinglish (regional understanding)

Intuition Hinglish mein samjho

Dekho, basic idea simple hai: jab current chalta hai, uske aas-paas magnetic field ban jaata hai jo wire ke chaaron taraf circle mein ghoomta hai. Direction nikalne ke liye right-hand grip rule lagao — angootha current ki direction mein, ungliyan jis taraf mudein wahi field. Field ki strength nikalne ke do hathiyaar hain: Biot–Savart law (jab koi symmetry nahi, integration karna padta hai) aur Ampère's law (jab symmetry ho, shortcut mil jaata hai).

Straight wire ke liye Ampère's law se ek circle Amperian loop banao, aur seedha aa jaata hai $B=\mu_0 I/(2\pi r)$ — yaani distance double karoge toh field aadha. Circular loop ke axis par symmetry nahi banti theek se, isliye Biot–Savart se integrate karna padta hai; off-axis components cancel ho jaate hain, sirf axial bachta hai, aur centre par $B=\mu_0 I/2a$ milta hai.

Solenoid matlab bahut saare loops ek line mein — inside field uniform ho jaata hai, $B=\mu_0 n I$ , aur yaad rakho yeh radius par depend nahi karta. Bahar field almost zero. Toroid matlab solenoid ko mod ke doughnut bana do — field poora andar hi trapped rehta hai, $B=\mu_0 N I/(2\pi r)$ , lekin yeh uniform nahi hai, inner edge par zyada strong hai kyunki $1/r$ hai.

Exam tip: chaaron formula ka pattern yaad rakho — straight aur toroid mein $2\pi r$ hai (circle wala), loop mein simple $2a$ , aur solenoid "naked" hai bina $r$ ke. Aur multi-turn coil ho toh $N$ se multiply karna mat bhoolna!

Magnetic field of straight wire, circular loop, solenoid, toroid

The two master laws (WHY they exist)

1. Straight Wire — Ampère's Law (HOW)

2. Circular Loop (on axis) — Biot–Savart (HOW)

3. Solenoid — Ampère's Law (HOW)

4. Toroid — Ampère's Law (HOW)

Forecast-then-Verify checkpoint

Flashcards

Connections

Concept Map

Hinglish (regional understanding)

Go deeper — visual, from zero

Test yourself — Electromagnetism

Connections