Intuition The one-line ideaA tiny piece of wire carrying current is a tiny magnet-maker . Each little segment d l ⃗ d\vec{l} d l d B ⃗ d\vec{B} d B circles around the segment (right-hand rule). To get the total field, you add up (integrate) the contributions of every segment along the wire.
Definition Biot–Savart LawThe magnetic field d B ⃗ d\vec{B} d B P P P I d l ⃗ I\,d\vec{l} I d l d B ⃗ = μ 0 4 π I d l ⃗ × r ^ r 2 d\vec{B} = \frac{\mu_0}{4\pi}\,\frac{I\,d\vec{l}\times \hat{r}}{r^2} d B = 4 π μ 0 r 2 I d l × r ^ r ⃗ \vec{r} r from the current element to the field point P P P r ^ = r ⃗ / r \hat{r}=\vec{r}/r r ^ = r / r μ 0 = 4 π × 10 − 7 T⋅m/A \mu_0 = 4\pi\times 10^{-7}\ \text{T·m/A} μ 0 = 4 π × 1 0 − 7 T⋅m/A
Key facts baked into this formula:
Direction: d B ⃗ d\vec{B} d B ⊥ \perp ⊥ both d l ⃗ d\vec{l} d l r ^ \hat{r} r ^ tangent to circles wrapping the current element.Magnitude: d B = μ 0 4 π I d l sin θ r 2 dB = \dfrac{\mu_0}{4\pi}\dfrac{I\,dl\,\sin\theta}{r^2} d B = 4 π μ 0 r 2 I d l sin θ θ \theta θ d l ⃗ d\vec{l} d l r ^ \hat{r} r ^ Falls off as 1 / r 2 1/r^2 1/ r 2 piece of current rather than a point charge.
Intuition Reading the formula like a sentence
I d l ⃗ I\,d\vec{l} I d l more current or longer segment ⇒ stronger field. A current element is the "source charge" equivalent.1 r 2 \dfrac{1}{r^2} r 2 1 sin θ \sin\theta sin θ zero field straight ahead of itself (θ = 0 \theta=0 θ = 0 sin θ = 0 \sin\theta=0 sin θ = 0 maximum field sideways (θ = 90 ∘ \theta=90^\circ θ = 9 0 ∘ perpendicular to their motion, never along it.× r ^ \times\hat r × r ^ circulating (curl-like) nature of magnetism: there are no magnetic "source points," only loops.
The bare formula is useless until we integrate it . Let's do the classic case from scratch.
Worked example Field at distancea a a Setup. Wire lies along the y y y I I I P P P a a a d l ⃗ = d y y ^ d\vec l = dy\,\hat{y} d l = d y y ^ y y y
Step 1 — geometry.
Let θ \theta θ d l ⃗ d\vec l d l r ^ \hat r r ^ r = a 2 + y 2 , sin θ = a a 2 + y 2 . r = \sqrt{a^2+y^2}, \qquad \sin\theta = \frac{a}{\sqrt{a^2+y^2}}. r = a 2 + y 2 , sin θ = a 2 + y 2 a . Why this step? The Biot–Savart magnitude needs r r r sin θ \sin\theta sin θ
Step 2 — write d B dB d B
= \frac{\mu_0 I}{4\pi}\frac{a\,dy}{(a^2+y^2)^{3/2}}.$$
*Why this step?* Substitute the geometry so everything is in $y$. All segments push $\vec B$ **out of the page** at $P$ (same direction), so we add magnitudes — no vector cancellation.
**Step 3 — integrate over the whole wire** ($y$ from $-\infty$ to $+\infty$):
$$B = \frac{\mu_0 I a}{4\pi}\int_{-\infty}^{\infty}\frac{dy}{(a^2+y^2)^{3/2}}.$$
The standard integral $\displaystyle\int\frac{dy}{(a^2+y^2)^{3/2}}=\frac{y}{a^2\sqrt{a^2+y^2}}$. Evaluated from $-\infty$ to $\infty$ gives $\dfrac{2}{a^2}$.
*Why this step?* Plugging limits: at $y\to+\infty$ the bracket $\to 1/a^2$; at $-\infty\to -1/a^2$; difference $=2/a^2$.
**Step 4 — result.**
$$\boxed{B = \frac{\mu_0 I}{2\pi a}}$$
*Why this matters:* This is the famous **infinite straight-wire field** — it falls as $1/a$ (not $1/r^2$!) because we summed infinitely many elements.
Worked example Field at the centre of a circular loop (radiusR R R Step 1. Every element d l ⃗ d\vec l d l perpendicular to r ^ \hat r r ^ θ = 90 ∘ \theta = 90^\circ θ = 9 0 ∘ sin θ = 1 \sin\theta=1 sin θ = 1 r = R r=R r = R Why? On a circle, the tangent is always perpendicular to the radius.
Step 2. d B = μ 0 I 4 π d l R 2 dB = \dfrac{\mu_0 I}{4\pi}\dfrac{dl}{R^2} d B = 4 π μ 0 I R 2 d l d B ⃗ d\vec B d B same way (along the axis through the centre).
Step 3. Integrate ∮ d l = 2 π R \oint dl = 2\pi R ∮ d l = 2 π R B = μ 0 I 4 π R 2 ( 2 π R ) = μ 0 I 2 R . B = \frac{\mu_0 I}{4\pi R^2}(2\pi R) = \boxed{\frac{\mu_0 I}{2R}}. B = 4 π R 2 μ 0 I ( 2 π R ) = 2 R μ 0 I .
Common mistake Common mistakes (Steel-manned)1. "Use r r r
Why it feels right: In many problems we measure from a fixed origin.
Fix: r ⃗ \vec r r always runs from the current element d l ⃗ d\vec l d l P P P , and it changes for every element.
2. "B = μ 0 I / ( 2 π a ) B = \mu_0 I/(2\pi a) B = μ 0 I / ( 2 π a )
Why it feels right: It's the formula everyone memorises.
Fix: That's only for an infinitely long straight wire. For finite wires use the sin θ 1 + sin θ 2 \sin\theta_1+\sin\theta_2 sin θ 1 + sin θ 2 μ 0 I / 2 R \mu_0 I/2R μ 0 I /2 R
3. Forgetting sin θ \sin\theta sin θ d l / r 2 dl/r^2 d l / r 2
Why it feels right: Looks like Coulomb's 1 / r 2 1/r^2 1/ r 2 Fix: Magnetism comes from a cross product ; the directional sin θ \sin\theta sin θ
4. Adding d B ⃗ d\vec B d B
Fix: Only add magnitudes when all d B ⃗ d\vec B d B P P P
Recall Predict before you compute
Double the current in the straight wire — what happens to B B B (Forecast…) → It doubles (B ∝ I B\propto I B ∝ I
Move twice as far from the straight wire — B B B Halves (B ∝ 1 / a B\propto 1/a B ∝ 1/ a
Move twice as far on a loop's axis ? → Falls faster than 1 / r 2 1/r^2 1/ r 2 μ 0 I R 2 2 ( R 2 + x 2 ) 3 / 2 \frac{\mu_0 I R^2}{2(R^2+x^2)^{3/2}} 2 ( R 2 + x 2 ) 3/2 μ 0 I R 2 ∝ 1 / x 3 \propto 1/x^3 ∝ 1/ x 3
Biot–Savart law (vector form) d B ⃗ = μ 0 4 π I d l ⃗ × r ^ r 2 d\vec B = \dfrac{\mu_0}{4\pi}\dfrac{I\,d\vec l\times\hat r}{r^2} d B = 4 π μ 0 r 2 I d l × r ^ What does r ⃗ \vec r r From the current element
d l ⃗ d\vec l d l to the field point
P P P .
Magnitude of d B dB d B d B = μ 0 4 π I d l sin θ r 2 dB=\dfrac{\mu_0}{4\pi}\dfrac{I\,dl\sin\theta}{r^2} d B = 4 π μ 0 r 2 I d l sin θ Why is field zero directly ahead of a current element? θ = 0 ⇒ sin θ = 0 \theta=0\Rightarrow\sin\theta=0 θ = 0 ⇒ sin θ = 0 (cross product vanishes).
Field of infinite straight wire B = μ 0 I 2 π a B=\dfrac{\mu_0 I}{2\pi a} B = 2 π a μ 0 I Field at centre of circular loop B = μ 0 I 2 R B=\dfrac{\mu_0 I}{2R} B = 2 R μ 0 I Finite straight wire field B = μ 0 I 4 π a ( sin θ 1 + sin θ 2 ) B=\dfrac{\mu_0 I}{4\pi a}(\sin\theta_1+\sin\theta_2) B = 4 π a μ 0 I ( sin θ 1 + sin θ 2 ) Value of μ 0 \mu_0 μ 0 4 π × 10 − 7 T⋅m/A 4\pi\times10^{-7}\ \text{T·m/A} 4 π × 1 0 − 7 T⋅m/A Direction of d B ⃗ d\vec B d B d l ⃗ d\vec l d l r ^ \hat r r ^ Perpendicular to both (right-hand rule), tangent to circles around the element.
On-axis field of a loop at distance x x x B = μ 0 I R 2 2 ( R 2 + x 2 ) 3 / 2 B=\dfrac{\mu_0 I R^2}{2(R^2+x^2)^{3/2}} B = 2 ( R 2 + x 2 ) 3/2 μ 0 I R 2
Recall Feynman: explain to a 12-year-oldImagine a hose that, instead of water, sprays out invisible "swirl." Each tiny bit of a wire with electricity flowing through it sprays swirl sideways — never straight forward and never straight back. The swirl makes little rings around the wire. Far away, the swirl is weaker. To find the total swirl at a spot, you add up the swirl from every tiny bit of the wire. That adding-up is the Biot–Savart law.
"I, d-L, cross-R, over R-squared" — chant the top of the fraction: Current, length-vector, cross r-hat. And: "Sideways max, straight-ahead zero" for the sin θ \sin\theta sin θ
Ampère's Law — easier shortcut for symmetric cases; gives same μ 0 I / 2 π a \mu_0 I/2\pi a μ 0 I /2 π a Magnetic Field of a Solenoid — built by integrating loop fields.Magnetic Dipole Moment — far-field of a loop is a dipole (∝ 1 / x 3 \propto 1/x^3 ∝ 1/ x 3 Coulomb's Law — the electric 1 / r 2 1/r^2 1/ r 2 Lorentz Force — what this B ⃗ \vec B B Right-Hand Rule — fixes all the directions here.
weakens field with distance
Biot-Savart law dB = mu0/4pi · I dl×r-hat/r^2
dB perpendicular to dl and r-hat
Field circles around segment
Integrate over whole wire
Field of long straight wire
Geometry r and sin theta in one variable
Intuition Hinglish mein samjhoDekho, Biot–Savart law ka idea simple hai: taar (wire) ka har chhota tukda jisme current beh rahi hai, woh apne aas-paas ek chhota magnetic field d B ⃗ d\vec B d B d B ⃗ = μ 0 4 π I d l ⃗ × r ^ r 2 d\vec B = \frac{\mu_0}{4\pi}\frac{I\,d\vec l\times \hat r}{r^2} d B = 4 π μ 0 r 2 I d l × r ^ r ⃗ \vec r r P P P side me, circle banake banta hai. Isliye sin θ \sin\theta sin θ θ = 0 \theta=0 θ = 0 θ = 90 ∘ \theta=90^\circ θ = 9 0 ∘
Ab sirf formula likhne se kaam nahi chalta — har element ka contribution add (integrate) karna padta hai. Long straight wire ke liye saare d B ⃗ d\vec B d B P P P B = μ 0 I 2 π a B=\frac{\mu_0 I}{2\pi a} B = 2 π a μ 0 I θ = 90 ∘ \theta=90^\circ θ = 9 0 ∘ B = μ 0 I 2 R B=\frac{\mu_0 I}{2R} B = 2 R μ 0 I
Ye important kyun hai? Kyunki yahi se motors, MRI, solenoid, sab ka magnetic field nikalta hai. Aur dhyaan rakho — μ 0 I 2 π a \frac{\mu_0 I}{2\pi a} 2 π a μ 0 I infinite straight wire ke liye hai; finite wire ke liye μ 0 I 4 π a ( sin θ 1 + sin θ 2 ) \frac{\mu_0 I}{4\pi a}(\sin\theta_1+\sin\theta_2) 4 π a μ 0 I ( sin θ 1 + sin θ 2 ) sin θ \sin\theta sin θ
