# Magnetic Field of a Current-Carrying Wire
*10 beats · narrated by Antoni · Canvas Tutor v0.5*
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## 1. Step 01 — Topic
> 🎙 *Here's a question that puzzled scientists for centuries: why does a wire carrying electric current affect a compass needle nearby? Today we'll uncover exactly what's going on — and derive the magnetic field around a long straight wire.*
> "Why does a wire carrying current deflect a compass — and how strong is the magnetic field it creates?"

## 2. Step 02 — Origin
> 🎙 *A moving charge creates a magnetic field around it. In a wire, billions of electrons drift along together — and their combined effect is a magnetic field that curls in circles around the wire. This is Oersted's discovery from 1820.*
Moving charges → magnetic field.In a wire, drifting electrons act together:
$$I = n q v_d A$$
The field curls in closed loops around the wire.
> ↗ Oersted (1820):current → compassdeflection. ⚡🧭
*[Diagram: 4 elements — baseline, text, concentricRings, vectorArrow 'B']*

## 3. Step 03 — Right-Hand Rule
> 🎙 *Here's the right-hand rule — the single most useful trick for magnetism. Point your right thumb in the direction the current flows. Your fingers curl in exactly the direction the magnetic field circles around the wire.*
Right-Hand Rule:① Point thumb → direction of I② Fingers curl → direction of B
*[Diagram: 6 elements — baseline, vectorArrow 'I', concentricRings, vectorArrow 'B', vectorArrow 'B', text]*
> ✓ Works every time —no memorising signs!

## 4. Step 04 — Setup
> 🎙 *Now let's set up the geometry carefully before we write any equations. We have a long straight wire carrying current I. We want to find the field at a perpendicular distance r from the wire's centre.*
Setup:• Long straight wire• Current I (amperes)• Point P at distance r from wire• Find B at P
*[Diagram: 6 elements — baseline, vectorArrow 'I', point 'P', dashedLine 'r', vectorArrow 'B = ?', text]*

## 5. Step 05 — Ampere's Law
> 🎙 *To derive the exact formula we use Ampere's Law — one of the four fundamental laws of electromagnetism. We draw an imaginary circular loop of radius r centred on the wire, called an Amperian loop.*
Ampere's Law (integral form):
$$\oint \vec{B} \cdot d\vec{\ell} \;=\; \mu_0 \, I_{\text{enc}}$$
Choose an Amperian loop: a circle of radius r around the wire. By symmetry, B is constant in magnitude and always tangent to this loop.
*[Diagram: 5 elements — point '⊙ I', concentricRings, dashedLine 'r', vectorArrow 'B', text]*
> ↗ μ₀ = 4π × 10⁻⁷ T·m/A(permeability of free space)

## 6. Step 06 — Derive
> 🎙 *Now let's work the integral. Because B has the same magnitude all the way around the loop, and it's always parallel to d-ell, the dot product is just B times the circumference two pi r. The enclosed current is simply I.*
Apply Ampere's Law step by step:
B constant & tangent → simplify:
$$B \oint d\ell \;=\; \mu_0 I$$
Circumference of loop = 2πr :
$$B \cdot 2\pi r \;=\; \mu_0 I$$
Solve for B :
$$B \;=\; \dfrac{\mu_0 I}{2\pi r}$$
*[Diagram: 5 elements — point '⊙ I', concentricRings, dashedLine 'r', vectorArrow 'dℓ', text]*

## 7. Step 07 — Result
> 🎙 *And there it is — the formula for the magnetic field of a long straight wire. B equals mu-naught I over two pi r. Let's box that result.*
Magnetic field of a long straight wireat perpendicular distance r :
> **Result:** $$B \;=\; \dfrac{\mu_0 \, I}{2\pi r}$$
> ↗ B in Tesla (T)I in Amperes (A)r in metres (m)

## 8. Step 08 — Verify
> 🎙 *Let's check the formula makes physical sense. If the current doubles, B doubles — that's proportional, makes sense. If r doubles, B halves — the field weakens as you move away. And if I equals zero, B equals zero — no current, no field. Perfect.*
**Verification:**
- $$I \to 2I \Rightarrow B \to 2B$$ — double the current → double the field ✓
- $$r \to 2r \Rightarrow B \to B/2$$ — twice as far → half the field ✓
- $$I = 0 \Rightarrow B = 0$$ — no current → no field ✓

## 9. Step 09 — Trap
> 🎙 *Here's the trap students fall into. The r in this formula is the perpendicular distance from the wire — not the distance along the wire, and not measured at some angle. Always draw a right-angle from the wire to your point.*
> ✗ ⚠ Common Mistake
r is the perpendicular distance from the wire.Not along the wire.Not at an angle.Always drop a right-angle to the wire first.
*[Diagram: 5 elements — baseline, text, point 'P', dashedLine 'r ✓', dashedLine '✗ not this']*

## 10. Step 10 — Try It
> 🎙 *Let's try a quick number: a wire carries 5 amperes of current. What's the magnetic field 10 centimetres away? Plug in: mu-naught is four-pi times ten to the minus seven, I is 5, r is 0.1 metres. You get B equals ten micro-tesla. Try it yourself!*
Quick calculation:Wire carries I = 5 A.Find B at r = 0.10 m.
**Student work:**
- $$B = \dfrac{\mu_0 I}{2\pi r}$$
- $$B = \dfrac{(4\pi \times 10^{-7})(5)}{2\pi (0.10)}$$
- $$B = \dfrac{4\pi \times 5 \times 10^{-7}}{2\pi \times 0.10}$$
- $$B = 1.0 \times 10^{-5} \text{ T} = 10\;\mu\text{T}$$
> **Result:** $$B = 10 \;\mu\text{T}$$
> ✓ Earth's field ≈ 50 μT —so this wire isweaker but measurable!