Magnetostatics & Ampèrian Loops: Master Notes

Chapter Overview

Magnetostatics governs steady electric currents and their induced magnetic fields. In IIT-JEE (Advanced), the most prestigious problems involve Ampère's Law with off-center cylindrical cavities, Helmholtz coil axial fields, and superposition of polygonal/circular current loops.

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1. Biot-Savart Law & Standard Configurations

$$d\overrightarrow{B}=\frac{{\mu }_0}{4\pi }\frac{Id\overrightarrow{\ell }\times \hat{r}}{r^2}=\frac{{\mu }_0}{4\pi }\frac{Id\overrightarrow{\ell }\times \overrightarrow{r}}{r^3}$$

Current Geometry	Magnetic Field Formula $B$	Notes
Straight Wire of Length $L$ at distance $d$	$B=\frac{{\mu }_{0}I}{4\pi d}(\sin {\theta }_1+\sin {\theta }_2)$	Infinite wire: $B=\frac{{\mu }_{0}I}{2\pi d}$
Circular Arc of Angle $\varphi$ at Center	$B=\frac{{\mu }_{0}I}{4\pi R}\varphi$	Full loop ($\varphi =2\pi$): $B=\frac{{\mu }_{0}I}{2R}$
Circular Ring on Axis at distance $x$	$B(x)=\frac{{\mu }_{0}I{R}^2}{2(R^2+x^2{)}^{3/2}}$	Far field: $B=\frac{{\mu }_0}{4\pi }\frac{2M}{x^3}$ ($M=I\pi R^2$)
Long Ideal Solenoid ($n$ turns/meter)	$B_{\text{inside}}={\mu }_{0}nI,B_{\text{end}}=\frac{1}{2}{\mu }_{0}nI$	Zero outside
Toroid ($N$ total turns)	$B=\frac{{\mu }_{0}NI}{2\pi r}$	Confined strictly inside core

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2. Ampère's Circuital Law & Magnetic Cavity Theorem

$${\oint }_{\mathcal{C}}\overrightarrow{B}\cdot d\overrightarrow{\ell }={\mu }_0{I}_{\text{enclosed}}$$

Multi-Mode DiagramMagnetostatics & Biot-Savart Magnetic Cavities

Option 1: Publication-Grade Scientific Vector SVG

Off-axis cylindrical magnetic cavity with uniform internal magnetic field B = (mu_0 / 2) * (J x a).

Cavity Magnetic Field: $\vec{B} = \frac{\mu_0}{2}(\vec{J} \times \vec{a}) = \text{Uniform}$

Option 2: Interactive 60 FPS Canvas Physics Engine

Simulate uniform magnetic field vector inside current-carrying cavity.

Current density J: 5 A/mm²

B_cavity ($B$):2.25 mT

Option 3: Interactive 3D Spatial Viewport

3D current cylinder with off-center bore.

+X

+Y

+Z

👆 Click and drag anywhere to rotate in 3D space

Azimuth Angle (ψ):35°

Elevation Angle (θ):-22°

Uniform Field:$\vec{B} = \text{const}$

Option 5: Reactive Parametric Physics with Live Calculus

Magnetic cavity field magnitude calculator.

Current Density J: 4 MA/m²

: 45

Field B ($B$):$2.52 mT$

Active Mode:Option 1 (Print & Derivation Vector SVG)Switch modes using the tabs above to explore different visualizations.

::: theorem 🌟 Magnetic Cavity Theorem For an infinitely long solid cylindrical conductor carrying uniform current density $\overrightarrow{J}=J\hat{k}$ having a parallel cylindrical cavity whose axis is displaced by vector $\overrightarrow{a}$:

$${\overrightarrow{B}}_{\text{cavity}}={\overrightarrow{B}}_{\text{solid}}-{\overrightarrow{B}}_{\text{removed}}=\frac{{\mu }_0}{2}(\overrightarrow{J}\times {\overrightarrow{r}}_1)-\frac{{\mu }_0}{2}(\overrightarrow{J}\times {\overrightarrow{r}}_2)=\frac{{\mu }_0}{2}\overrightarrow{J}\times ({\overrightarrow{r}}_1-{\overrightarrow{r}}_2)$$$$\boxed{{\displaystyle {\overrightarrow{B}}_{\text{cavity}}=\frac{{\mu }_0}{2}(\overrightarrow{J}\times \overrightarrow{a})=\text{constant in magnitude and direction everywhere inside!}}}$$

:::

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3. Magnetic Dipole Moment & Torque

$$\overrightarrow{M}=NI\overrightarrow{A},\overrightarrow{\tau }=\overrightarrow{M}\times \overrightarrow{B},U=-\overrightarrow{M}\cdot \overrightarrow{B}$$

• Gyromagnetic Ratio: For any uniformly charged rotating rigid body with total charge $Q$ and mass $M$:$$\boxed{{\displaystyle \frac{\overrightarrow{M}}{\overrightarrow{L}}=\frac{Q}{2M}}}$$(This universal ratio holds for a spinning sphere, ring, disk, or cylinder!)

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4. Authentic Previous Years Questions (PYQs)

PYQ 1: JEE Advanced 2022 (Paper 1) — Cylindrical Cavity Magnetic Field

Question:
A long cylindrical conductor of radius $R$ carries a uniform current density $J$ along its axis. A cylindrical hole of radius $R/2$ is drilled parallel to the axis, with its center displaced by distance $d=R/2$ along the $x$-axis ($\overrightarrow{a}=\frac{R}{2}\hat{i}$). Find the magnetic field vector $\overrightarrow{B}$ at any point inside the hole.

Step-by-Step Solution:

Using the Magnetic Cavity Theorem:

$$\overrightarrow{J}=J\hat{k},\overrightarrow{a}=\frac{R}{2}\hat{i}$$$$\overrightarrow{B}=\frac{{\mu }_0}{2}(\overrightarrow{J}\times \overrightarrow{a})=\frac{{\mu }_0}{2}(J\hat{k}\times \frac{R}{2}\hat{i})=\frac{{\mu }_{0}JR}{4}(\hat{k}\times \hat{i})=\boxed{{\displaystyle \frac{{\mu }_{0}JR}{4}\hat{j}}}$$

(The field is completely uniform and directed along the $+y$ axis!)

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PYQ 2: JEE Advanced 2020 — Spinning Charged Sphere Dipole Moment

Question:
A thin spherical shell of radius $R$ and uniform surface charge density $\sigma$ (total charge $Q$) rotates with constant angular velocity $\omega$ about its diameter. Calculate the magnetic dipole moment $M$ of the spinning shell.

Step-by-Step Solution:

1. Slice the spherical shell into horizontal rings at polar angle $\theta \in [0,\pi ]$.
2. The radius of a ring at angle $\theta$ is $r(\theta )=R\sin \theta$.
3. Area of ring: $dA=2\pi R\sin \theta (Rd\theta )=2\pi R^2\sin \theta d\theta$.
4. Charge on ring: $dq=\sigma dA=\left(\frac{Q}{4\pi R^2}\right)2\pi R^2\sin \theta d\theta =\frac{Q}{2}\sin \theta d\theta$.
5. Current of ring: $dI=\frac{dq}{T}=dq\frac{\omega }{2\pi }=\frac{Q\omega }{4\pi }\sin \theta d\theta$.
6. Magnetic moment of differential ring:$$dM=dI\cdot \text{Area}=(\frac{Q\omega }{4\pi }\sin \theta d\theta )(\pi R^2{\sin }^2\theta )=\frac{1}{4}Q\omega R^2{\sin }^3\theta d\theta$$
7. Integrate from $\theta =0$ to $\pi$:$$M=\frac{1}{4}Q\omega R^2{\int }_0^{\pi }{\sin }^3\theta d\theta =\frac{1}{4}Q\omega R^2\left(\frac{4}{3}\right)=\boxed{{\displaystyle \frac{1}{3}Q\omega R^2}}$$

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5. High-Yield Formula Sheet

Entity	Formula	Notes
Wire Field	$B=\frac{{\mu }_{0}I}{4\pi d}(\sin {\theta }_1+\sin {\theta }_2)$	Finite straight segment
Ring on Axis	$B(x)=\frac{{\mu }_{0}I{R}^2}{2(R^2+x^2{)}^{3/2}}$	Max at center $x=0$
Cavity Field	$\overrightarrow{B}=\frac{{\mu }_0}{2}(\overrightarrow{J}\times \overrightarrow{a})$	Uniform in cavity
Gyromagnetic Ratio	$\frac{M}{L}=\frac{Q}{2M}$	Uniform $Q/M$ distribution
Spinning Shell Dipole	$M=\frac{1}{3}Q\omega R^2$	Spinning spherical shell