Comprehensive Study Notes on Magnetostatics and Ampere's Law Applications

Magnetic Field Due to an Infinitely Long Sheet of Current

• Concept Overview: The magnetic field produced by an infinitely long thin sheet carrying a uniform surface current is evaluated using Ampere’s Circuital Law.
• Mathematical Formulation:
  • Let the sheet have a surface current density $\lambda$.
  • Using an Amperian loop $ABCD$ of length $L$:         $\oint \mathbf{B} ∙ d\mathbf{l} = \mu_0 I_{in}$$\int_{AB} \mathbf{B} ∙ d\mathbf{l} + \int_{BC} \mathbf{B} ∙ d\mathbf{l} + \int_{CD} \mathbf{B} ∙ d\mathbf{l} + \int_{DA} \mathbf{B} ∙ d\mathbf{l} = \mu_0 (\lambda L)$
  • Since the field is parallel to segments $AB$ and $CD$ and perpendicular to $BC$ and $DA$:         $B(L) + 0 + B(L) + 0 = \mu_0 (\lambda L)$$2 B L = \mu_0 \lambda L$
  • Final Expression: $B = \frac{\mu_0 \lambda}{2}$
• Thick Sheet Case: For a sheet of thickness $d$ with a uniform volume current density $j$:
  • Inside the sheet ($x < d$): The field increases linearly as $B = \mu_0 j x$.
  • Outside the sheet ($x > d$): The field is constant, equivalent to a thin sheet where $\lambda = j d$, giving $B = \frac{\mu_0 j d}{2}$.

Magnetic Field Due to a Long Solenoid

• Ideal Solenoid Properties:
  • The magnetic field inside the solenoid ($B_{inside}$) is uniform near the middle portion.
  • The magnetic field just outside the solenoid ($B_{outside}$) is approximately zero.
• Derivation using Ampere's Law:
  • Choosing a rectangular Amperian loop $ABCD$ where side $AB$ (of length $L$) is inside and parallel to the axis, and side $CD$ is outside.
  • $\oint \mathbf{B} ∙ d\mathbf{l} = \mu_0 I_{in}$
  • $B ∙ L + 0 + 0 + 0 = \mu_0 (NI)$, where $N$ is the number of turns enclosed by the loop.
  • $B = \mu_0 \left(\frac{N}{L}\right) I$
  • Final Expression: $B = \mu_0 n I$, where $n$ is the number of turns per unit length.

Magnetic Field Due to a Toroid

• Configuration: A toroid is essentially a solenoid bent into a circular shape.
• Field Distribution:
  • $B = 0$ in the open space interior to the toroid.
  • $B = 0$ in the space exterior to the toroid.
  • The field exists only within the core of the toroid.
• Derivation:
  • Using a circular Amperian loop of radius $r$ within the toroid core:         $\oint \mathbf{B} ∙ d\mathbf{l} = \mu_0 I_{total}$$B (2 \pi r) = \mu_0 (N I)$
  • Final Expression: $B = \frac{\mu_0 N I}{2 \pi r}$
  • If defined by turns per unit length $n = \frac{N}{2 \pi r}$, then $B = \mu_0 n I$.

Line Integrals of Magnetic Induction Fields

• JEE Advanced 2024 Problem: An infinitely long wire on the $z$-axis carries a current $I$ in the $+z$-direction. The line integral $\int \mathbf{B} ∙ d\mathbf{l}$ along a straight line from point $P_1(-\sqrt{3}a, a, 0)$ to $P_2(a, a, 0)$ is required.
  • The angle subtended by the line at the wire is calculated.
  • $\theta_1 = \tan^{-1}\left(\frac{a}{-\sqrt{3}a}\right) = 150^{\circ}$
  • $\theta_2 = \tan^{-1}\left(\frac{a}{a}\right) = 45^{\circ}$
  • Difference in angles ($\Delta \phi$) is calculated based on the geometry (e.g., $60^{\circ} + 45^{\circ} = 105^{\circ}$, leading to specific fractional values like $7/24$ or $1/8$).
  • Final Result: $\frac{7 \mu_0 I}{24}$.
• Closed Path Integral Example: For a closed path $ABCDEFA$ enclosing currents $I_2$ (downward) and $I_3$ (upward):
  • $I_{enclosed} = I_3 - I_2$
  • $\oint \mathbf{B} ∙ d\mathbf{l} = \mu_0 (I_3 - I_2)$.
• Straight Line Integral Subtending Angle $\theta$: For a wire into the plane and a line $AB$ subtending angle $\theta$ at the wire:
  • The line integral along the segment $AB$ is $\int_A^B \mathbf{B} ∙ d\mathbf{l} = \frac{\mu_0 I \theta}{2 \pi}$.

Non-Uniform Current Density in Cylinders

• Problem Scenario: A long cylindrical region of radius $a$ carries a current with density $J = J_0 (1 - \frac{r}{a})$.
• Mean Current Density:
  • Total Current $I = \int_0^a J (2 \pi r) dr = 2 \pi J_0 \int_0^a (r - \frac{r^2}{a}) dr = 2 \pi J_0 [\frac{a^2}{2} - \frac{a^2}{3}] = \frac{\pi J_0 a^2}{3}$.
  • $J_{mean} = \frac{I}{\pi a^2} = \frac{J_0}{3}$.
• Magnetic Field Variation ($r \le a$):
  • Applying Ampere's Law at radius $r$:         $B (2 \pi r) = \mu_0 \int_0^r J_0 (1 - \frac{r'}{a}) (2 \pi r') dr' = \mu_0 2 \pi J_0 [\frac{r^2}{2} - \frac{r^3}{3a}]$
  • $B(r) = \mu_0 J_0 [\frac{r}{2} - \frac{r^2}{3a}]$.
  • The field follows a downward parabola. The maximum field occurs at $\frac{dB}{dr} = 0 \rightarrow \frac{1}{2} - \frac{2r}{3a} = 0 \rightarrow r = \frac{3a}{4}$.

Cylindrical Conductors with Cavities

• Basic Principle: The magnetic field inside a cavity is determined by superimposing the field of a solid conductor and the field of a conductor with the opposite current density filling the cavity space.
• Uniform Internal Field: It is shown that if the cavity axis is parallel to the cylinder axis and displaced by distance $l$, the field inside the cavity is uniform:
  • $\mathbf{B}_{cavity} = \frac{\mu_0}{2} (\mathbf{j} \times \mathbf{l})$.
• Case Study: Cylinder Diameter $2a$ with Cavity Diameter $a$:
  • Current density $j$ is uniform.
  • At point $P$ (external to center): $B_P = B_{complete} - B_{cavity}$.
  • Using the provided parameters, the magnitude of the field is given by $\frac{N}{12} \mu_0 a J$. If the calculation results in a specific ratio like $5/12$, then $N = 5$.
• Case Study: Two Cavities of Diameter $R$ in Cylinder of Radius $R$:
  • Radius of conductor is $R$. Cavities have diameter $R$ (radius $R/2$).
  • Field at distance $r = 2R$ from the conductor axis:         $B_P = B_{complete} - 2 B_{cavity} \cos(\theta)$$B_P = \frac{\mu_0 I}{2 \pi (2R)} [\text{terms related to geometry}]$
  • Calculated result provided: $B = \frac{9 \mu_0 I}{34 \pi R}$.

Special Integral Problems

• Integral Along Coil Axis: For a circular coil of $N$ turns, the integral $\int_{-\infty}^{+\infty} B \, dx$ along its axis is found to be:
  • $\int_{-\infty}^{+\infty} \frac{\mu_0 N I R^2}{2(R^2 + x^2)^{3/2}} dx$
  • Final Result: $\int \mathbf{B} ∙ d\mathbf{x} = \mu_0 N I$.
• Radial Current on a Plane: A straight wire carries current $I$ to point $O$ and then spreads radially over an infinite plane.
  • In the half-space with the straight wire: $B = \frac{\mu_0 I}{2 \pi r}$.
  • In the half-space without the wire: $B = 0$.
• Helical Strip Solenoid: A strip of width $h$ is wound around a radius $R$ to form a solenoid.
  • If the winding has an angle $\alpha$:         $B_{inside} = \mu_0 n I$ where $n = \frac{\cos(\alpha)}{h}$.
  • $B_{outside} = \frac{\mu_0 I \sin(\alpha)}{2 \pi r}$.

Dimensions and General Physics

• Dimensions of $\sqrt{\mu/\epsilon}$: The term $\sqrt{\frac{\mu}{\epsilon}}$ has the same dimensions as Resistance (it is known as the intrinsic impedance of free space, approximately $377 \, \Omega$).
• Arc-Segment Magnetic Field: A current $I$ flows in a path with eight alternating arcs of radii $r$ and $2r$.
  • Contribution of each arc at the center $P$: $d B = \frac{\mu_0 I}{4 \pi R} \theta$.
  • Total path subtends $360^{\circ} (2\pi)$. Since arcs alternate, each segment subtends $\frac{2 \pi}{8} = \frac{\pi}{4}$.
  • Summing contributions: $B = 4 \times \frac{\mu_0 I (\pi/4)}{4 \pi r} + 4 \times \frac{\mu_0 I (\pi/4)}{4 \pi (2r)} = \frac{\mu_0 I}{4r} + \frac{\mu_0 I}{8r} = \frac{3 \mu_0 I}{8r}$.
  • Direction is determined by the Right Hand Rule (e.g., inward or outward depending on flow).

Questions & Discussion

• Observer Facing West: One loop has clockwise current and the other has counter-clockwise current for an observer facing west.
  • The net magnetic field at their common center results from the subtraction or addition of individual fields based on the observer's perspective.
  • Specific Answer: $5 \times 10^{-4} \, T$ towards east or west depending on the specific current magnitudes and loop sizes (Options provided: $5 \times 10^{-4} \, T$ or $13 \times 10^{-4} \, T$).