2A · Biot–Savart — Wire + Loop (Full Solutions)

Biot–Savart law, straight-wire derivation with 2023 numeric, loop-axis derivation. Step-by-step, SI units.

Biot–Savart law is the magnetic version of Coulomb law. A tiny piece of current $I,d\vec{l}$ makes a tiny magnetic field $d\vec{B}$. We add (integrate) all such pieces to get the total field of any wire or loop.

Formula 1. $d\vec{B} = \dfrac{\mu_0}{4\pi}\,\dfrac{I\,d\vec{l}\times\hat{r}}{r^2}$, where $d\vec{l}$ points along the current, $\hat{r}$ points from the piece to the field point, and $r$ is the distance between them.

Direction rule. Grip the wire with the right hand, thumb along the current. The fingers curl along $\vec{B}$ (concentric circles round the wire). The cross product gives the sense without ambiguity.

Unit check. $[\mu_0 I,dl/r^2] = (\text{N/A}^2)(\text{A}\cdot\text{m})/\text{m}^2 = \text{N/(A·m)} = \text{tesla (T)}$. Correct.

Current loop of radius R, axial point at distance x, current I

Fig 1. Loop of radius $R$ carrying current $I$; field point on the axis at distance $x$. Diagram matches the derivation of Eq. (3) below.

Straight wire — field at distance r

Take an infinite straight wire along the $z$-axis. A point P sits at perpendicular distance $r$. Each current element contributes a piece of field. After integration (substitute $l = r\tan\phi$) the field forms complete loops around the wire with magnitude

Formula 2. $B = \dfrac{\mu_0 I}{2\pi r}$ — tangent to circles around the wire.

Circular loop — field on the axis

Take a loop of radius $R$ carrying current $I$. Look at a point P on the axis at distance $x$ from the centre. By symmetry the perpendicular components cancel in pairs. Only the axial component survives. Integrating round the loop gives

Formula 3. $B(x) = \dfrac{\mu_0 I R^2}{2(R^2+x^2)^{3/2}}$, directed along the axis by the right-hand rule.

At the centre ($x=0$) this reduces to $B = \mu_0 I/2R$. Far away ($x \gg R$) it falls as $1/x^3$, the dipole behaviour.

Solved PYQs

Q1201820192022×3

State Biot–Savart's law in magnetostatics.

Recall

The vector form is $d\vec{B} = (\mu_0/4\pi)\,I\,d\vec{l}\times\hat{r}/r^2$, magnitude $dB = (\mu_0/4\pi)\,I\,dl\,\sin\theta/r^2$, direction by right-hand rule.

Steps

Write the law in vector form: $d\vec{B} = (\mu_0/4\pi)\,I\,d\vec{l}\times\hat{r}/r^2$.
Define every symbol: $I$ in A, $d\vec{l}$ in m along the current, $r$ in m, $\hat{r}$ from piece to field point.
Write the magnitude form: $dB = (\mu_0/4\pi)\,I\,dl\,\sin\theta/r^2$, where $\theta$ is the angle between $d\vec{l}$ and $\hat{r}$.
State the direction: $\vec{B}$ is perpendicular to the plane of $d\vec{l}$ and $\hat{r}$; right-hand rule fixes the sense.
Total field of a wire or loop is the integral: $\vec{B} = \int d\vec{B}$.

Answer

$\boxed{\,d\vec{B} = \dfrac{\mu_0}{4\pi}\,\dfrac{I\,d\vec{l}\times\hat{r}}{r^2}\,}$ — vector form, direction by right-hand rule. SI unit: tesla (T).

Exam tip

Write the vector form first, then the $\sin\theta$ magnitude form — both versions earn marks.

Q2201920222023×3

Find the magnetic induction at distance $r$ from a long straight conductor carrying current $I$. (2023 numeric: $I = 1.5$ A, $r = 3$ cm.)

Recall

Apply Formula 1 and integrate along the wire. The result is $B = \mu_0 I/2\pi r$, tangent to circles around the wire.

Steps

Set geometry: wire along $z$, point P at perpendicular distance $r$, a piece at angle gives $dB = (\mu_0/4\pi)\,I\,dl\,\sin\theta/r'^2$ with $r'$ the slant distance.
Change variable to angle $\phi$ using $l = r\tan\phi$, so $dl\,\sin\theta/r'^2 = \cos\phi\,d\phi/r$.
Integrate $\phi$ from $-\pi/2$ to $+\pi/2$ (infinite wire): $B = (\mu_0 I/4\pi r)\int_{-\pi/2}^{\pi/2}\cos\phi\,d\phi = (\mu_0 I/4\pi r)\times 2$.
Apply the 2023 numbers: $\mu_0/2\pi = 2\times 10^{-7}$ (T·m/A), $I = 1.5$ A, $r = 0.03$ m.
Compute: $B = 2\times 10^{-7}\times 1.5/0.03 = 3.0\times 10^{-7}/0.03 = 1.0\times 10^{-5}$ T.

Answer

General: $\boxed{\,B = \dfrac{\mu_0 I}{2\pi r}\,}$ tangent to circles around the wire. Numeric (2023): $\boxed{\,B = 1.0\times 10^{-5}\,\text{T} = 10\,\mu\text{T}\,}$.

Exam tip

Convert 3 cm → 0.03 m first. Forgetting this is the most common zero-mark error.

Q320182023×2

Applying Biot–Savart's law, derive $\vec{B}$ at a point on the axis of a current-carrying circular loop.

Recall

Each piece $d\vec{l}$ is perpendicular to $\hat{r}$, so $\sin\theta = 1$. Perpendicular components from opposite pieces cancel in pairs — only the axial component survives.

Steps

Geometry: loop radius $R$, current $I$, point P on the axis at distance $x$ from the centre; slant distance $\rho = \sqrt{R^2+x^2}$.
Field of one piece: $dB = (\mu_0/4\pi)\,I\,dl/\rho^2$ because $d\vec{l}\perp\hat{r}$.
Keep the axial part: $dB_x = dB\cos\alpha$ with $\cos\alpha = R/\rho = R/\sqrt{R^2+x^2}$ (perpendicular parts cancel by symmetry).
Integrate round the loop: $\int dl = 2\pi R$, so $B = (\mu_0/4\pi)(I/\rho^2)(R/\rho)(2\pi R) = \mu_0 I R^2/2\rho^3$.
Centre case $x=0$: substitute to get $B(0) = \mu_0 I/2R$.

Answer

$\boxed{\,B(x) = \dfrac{\mu_0 I R^2}{2(R^2+x^2)^{3/2}}\,}$ along the loop axis (sense by right-hand rule).

Exam tip

Draw the loop with $R$, $x$, $I$ labelled as in Fig 1, then write Step 3 before integrating.

Reference. Brij Lal & Subrahmanyam Ch-7; H.C. Verma Vol-1 Ch-35; S.L. Arora Vol-1 Ch-4.