Moving Charges & Magnetism Derivations · Class XII

01

Biot-Savart's Law

Magnetic field due to a current element

The magnetic field $d\vec{B}$ at a point P due to a tiny current element $Id\vec{l}$ is perpendicular to both $d\vec{l}$ and $\hat{r}$, and falls off as $1/r^2$.

Fig 1.1 — Current element $Id\vec{l}$ at O; $\vec{r}$ to field point P; $d\vec{B}$ points out of page

Derivation

Consider a conductor carrying current $I$. Take a tiny element $d\vec{l}$ at origin. Let $\vec{r}$ be the position vector from this element to point P, making angle $\theta$ with $d\vec{l}$.

①

From experiment, the magnitude of $d\vec{B}$ is: $$dB \propto \frac{I\,dl\,\sin\theta}{r^2}$$

②

Introducing proportionality constant $\dfrac{\mu_0}{4\pi}$ (in SI): $$dB = \frac{\mu_0}{4\pi}\cdot\frac{I\,dl\,\sin\theta}{r^2}$$

③

In vector form, using cross product ($d\vec{B} \perp d\vec{l}$ and $\perp \hat{r}$):

★ Result $$d\vec{B} = \frac{\mu_0}{4\pi}\cdot\frac{I\,(d\vec{l}\times\hat{r})}{r^2} = \frac{\mu_0}{4\pi}\cdot\frac{I\,(d\vec{l}\times\vec{r})}{r^3}$$

Direction: Right-hand rule — curl fingers from $d\vec{l}$ to $\hat{r}$; thumb gives $d\vec{B}$. | $\mu_0$ $= 4\pi\times10^{-7}\;\text{T\,m\,A}^{-1}$ (permeability of free space).

02

Magnetic Field — Circular Coil

At Centre & At Axis

A circular loop of radius $R$ carrying current $I$ — we apply Biot-Savart and integrate around the loop. Two cases: centre of the loop, and a point P on the axis at distance $x$.

Case I — At the Centre

Fig 2.1 — Every element $dl$ is perpendicular to $r = R$, so $\sin\theta = 1$ always

Derivation · Centre

For any element $dl$ on the loop, $r = R$ and $\theta = 90°$ (since $d\vec{l} \perp \vec{r}$ always at the centre). So $\sin 90° = 1$.

①

Field due to one element by Biot-Savart: $$dB = \frac{\mu_0}{4\pi}\cdot\frac{I\,dl}{R^2}$$

②

By symmetry all $d\vec{B}$ point in the same direction (along axis). Integrate over the full loop (circumference $= 2\pi R$): $$B = \oint dB = \frac{\mu_0 I}{4\pi R^2}\oint dl = \frac{\mu_0 I}{4\pi R^2}\cdot 2\pi R$$

★ Result $$B = \frac{\mu_0 I}{2R}$$ For N turns: $B = \dfrac{\mu_0 N I}{2R}$

Case II — At a Point on the Axis

Fig 2.2 — Perpendicular components of $dB$ cancel in pairs; only axial components add up

Derivation · Axis

Let P be at distance $x$ from centre. Distance from each element to P is $r = \sqrt{R^2+x^2}$. Each $dB$ is perpendicular to $\vec{r}$, so it has components — axial (along axis) and perpendicular. By symmetry, perpendicular parts cancel.

①

Magnitude of $dB$ (since $\theta = 90°$, $d\vec{l} \perp \vec{r}$): $$dB = \frac{\mu_0}{4\pi}\cdot\frac{I\,dl}{R^2+x^2}$$

②

Axial component of each $dB$ (angle $\phi$ with axis, where $\cos\phi = R/r$): $$dB_{\text{axial}} = dB\cdot\frac{R}{\sqrt{R^2+x^2}}$$

③

Integrate over full loop ($\oint dl = 2\pi R$): $$B = \frac{\mu_0 I}{4\pi}\cdot\frac{2\pi R}{(R^2+x^2)}\cdot\frac{R}{\sqrt{R^2+x^2}}$$

★ Result $$B = \frac{\mu_0 I R^2}{2(R^2+x^2)^{3/2}}$$ At centre $(x=0)$: reduces to $B = \dfrac{\mu_0 I}{2R}$ ✓

Special case: When $x \gg R$, $(R^2+x^2)^{3/2} \approx x^3$, so $B \approx \dfrac{\mu_0 IR^2}{2x^3}$ — like a magnetic dipole.

03

Ampere's Circuital Law

∮ B·dl = μ₀ I_enc

The line integral of magnetic field around any closed loop (Amperian loop) equals $\mu_0$ times the total current enclosed by that loop.

Fig 3.1 — Circular Amperian loop of radius $r$ around a long straight wire; $\vec{B} \parallel d\vec{l}$ everywhere on loop

Statement & Derivation

Statement: $\oint_C \vec{B}\cdot d\vec{l} = \mu_0 I_{\text{enc}}$, where $I_{\text{enc}}$ is the net current threading through the loop.

①

From Biot-Savart, field at distance $r$ from an infinite straight wire is: $$B = \frac{\mu_0 I}{2\pi r}$$ Direction: tangential (by right-hand rule).

②

Choose an Amperian circle of radius $r$ around wire. Since $\vec{B} \parallel d\vec{l}$ at every point on the loop: $$\oint \vec{B}\cdot d\vec{l} = B\oint dl = B\cdot 2\pi r$$

③

Substituting $B = \dfrac{\mu_0 I}{2\pi r}$: $$\oint \vec{B}\cdot d\vec{l} = \frac{\mu_0 I}{2\pi r}\cdot 2\pi r$$

★ Result $$\oint_C \vec{B}\cdot d\vec{l} = \mu_0\, I_{\text{enc}}$$

Key insight: The result is independent of the shape of the Amperian loop and only depends on the enclosed current — not on $r$. Currents outside the loop contribute zero net integral.

Application — Magnetic Field inside a Solenoid

Quick Derivation

Take a rectangular Amperian loop ABCD with AB of length $l$ inside the solenoid (field $= B$) and CD outside (field $= 0$). Sides BC and AD contribute nothing ($\vec{B}\perp d\vec{l}$).

$$\oint \vec{B}\cdot d\vec{l} = B\cdot l + 0 + 0 + 0 = Bl$$

If $n$ = turns per unit length, enclosed current $= nIl$. By Ampere's Law:

★ Result $$B = \mu_0 n I$$ Valid deep inside; field is uniform & parallel to axis

04

Lorentz Force (Force on Moving Charge)

F = q(v × B)

A charge $q$ moving with velocity $\vec{v}$ in a magnetic field $\vec{B}$ experiences a force perpendicular to both — it cannot do work on the charge.

Fig 4.1 — $+q$ moves rightward in $\vec{B}$ (into page); force is upward by right-hand rule

Derivation

Consider a conductor of length $l$, cross-section $A$, carrying current $I$ in field $\vec{B}$. Let $n$ = free electron density, each moving with drift velocity $\vec{v}_d$.

①

Total charge in the conductor: $q = nAl\cdot e$.

②

Force on this conductor in $\vec{B}$: $$\vec{F} = I(\vec{l}\times\vec{B})$$ But $I = nAev_d$ and $q = nAle$, so $I\vec{l} = nAle\cdot\vec{v}_d = q\vec{v}_d$.

③

Force on a single charge $q$ moving with velocity $\vec{v}$:

★ Lorentz Force $$\vec{F} = q(\vec{v}\times\vec{B})$$ $$|\vec{F}| = qvB\sin\theta$$

Direction: $\vec{F} \perp \vec{v}$ always, so the magnetic force does no work — it changes direction, not speed.

Circular motion: If $\vec{v}\perp\vec{B}$, the charge moves in a circle: $r = \dfrac{mv}{qB}$.

05

Torque on a Current Loop

τ = NIAB sinθ = MB sinθ

A rectangular current loop placed in a uniform magnetic field experiences zero net force but a net torque that tends to align the loop's magnetic moment with $\vec{B}$.

Fig 5.1 — Forces $F_1$ and $F_2$ on AB & DC form a couple; AD and BC forces are collinear and cancel

Derivation

Let the rectangular loop have sides $a$ (perpendicular to $\vec{B}$) and $b$ (parallel to $\vec{B}$), carrying current $I$, tilted at angle $\theta$ with $\vec{B}$.

①

Forces on sides AD and BC (parallel to $\vec{B}$): $F = BIb\sin0° = 0$. They cancel. No contribution.

②

Forces on sides AB and DC (perpendicular to $\vec{B}$): $$F_1 = F_2 = BIb$$ They are equal, opposite, and not collinear — forming a couple.

③

Perpendicular distance between $F_1$ and $F_2$ is $a\sin\theta$. Torque of the couple: $$\tau = F_1\cdot a\sin\theta = BIb\cdot a\sin\theta = BIA\sin\theta$$ where $A = ab$ is the area of the loop.

★ Result $$\tau = NIAB\sin\theta = MB\sin\theta$$ where $M = NIA$ is the magnetic dipole moment of the loop

Vector form: $\vec{\tau} = \vec{M}\times\vec{B}$ — $\tau$ is max when $\theta = 90°$ (loop parallel to $\vec{B}$), zero when $\theta = 0°$ (stable equilibrium).

Analogy: Just like electric dipole in $\vec{E}$, current loop acts as a magnetic dipole in $\vec{B}$. $M = NIA$ is the magnetic moment.