Unit 03: Magnetic Effects of Current • Chapter 04[cite: 12, 13]
Ampere's Circuital Law & Straight Solenoid
Closed-loop line integrals, Amperian loops, infinite straight wire field, and interior uniform field of a long solenoid.
Electrostatics mein jab complex charge distribution ka field nikalna hota tha, toh hum Coulomb's Law ke lambe integration se bachne ke liye Gauss's Law (surface integral $\oint \vec{E}\cdot d\vec{A}$) use karte the. Theek waise hi, magnetostatics mein Biot-Savart Law ke lambe calculus se bachne ke liye hamare paas Ampere's Circuital Law hai! Farq sirf itna hai ki Gauss's law me ek 3D closed surface charge ko qaid karti hai, jabki Ampere's law me ek 2D closed loop (Amperian loop) wire ke current ko lasso ki tarah gher leta hai. Agar loop ke andar se koi net current pass nahi ho raha, toh chahe bahar kitna bhi current ho, loop ke along net circulation hamesha zero aayega!
1. Ampere's Circuital Law (ACL)
Line integral of magnetic field $\vec{B}$ around any closed path equals $\mu_0$ times the total threading current:
Right-hand rule sign convention: Curl fingers along loop integration direction; thumb points along positive current direction.
2. Straight Wire & Long Solenoid Field
Core applications derived via Amperian symmetry:
Where $n = N/L$ is the number of turns per unit length[cite: 13]. At extreme ends of a finite solenoid: $B_{\text{end}} \approx \frac{1}{2}\mu_0 n I$.
Magnetic Field of an Infinitely Long Straight Wire using ACL[cite: 13]
1. Consider an infinitely long straight wire carrying steady current $I$. By cylindrical symmetry, magnetic field lines are concentric circles in planes perpendicular to the wire[cite: 13].
2. Construct a circular Amperian loop of radius $r$ centered on the wire. At every point on this loop, $\vec{B}$ is tangential to the circle, and its magnitude $B$ is constant due to radial symmetry.
3. Tangential element vector $d\vec{l}$ is parallel to $\vec{B}$ everywhere ($\theta = 0^\circ \implies \vec{B}\cdot d\vec{l} = B\,dl\cos 0^\circ = B\,dl$):
4. Total current enclosed by the Amperian loop is $I_{\text{enc}} = I$. Applying Ampere's Circuital Law[cite: 13]:
5. Solving for magnetic field magnitude[cite: 13]:
Magnetic Field Inside a Long Straight Solenoid ($B = \mu_0 n I$)[cite: 13]
1. Consider a long tightly wound solenoid having $n$ turns per unit length carrying current $I$[cite: 13]. Inside the solenoid, field lines are parallel and dense (strong uniform field); outside, field lines spread out over vast volume and are nearly zero ($\vec{B}_{\text{out}} \approx 0$).
2. Construct a rectangular Amperian loop $abcd$ of length $L$ and width $w$, where side $ab$ lies parallel to the axis inside the solenoid, sides $bc$ and $da$ are perpendicular to the axis, and side $cd$ lies completely outside.
3. Evaluate the closed line integral across all 4 segments:
4. Individual segment breakdown:
• Along $ab$ (inside): $\vec{B} \parallel d\vec{l} \implies \int_a^b B\,dl = B L$.
• Along $bc$ and $da$: $\vec{B} \perp d\vec{l} \implies \cos 90^\circ = 0 \implies \int_b^c = \int_d^a = 0$.
• Along $cd$ (outside): $\vec{B} \approx 0 \implies \int_c^d = 0$.
5. Number of turns enclosed by the loop of length $L$ is $N_{\text{enc}} = n L$. Total enclosed current $I_{\text{enc}} = (n L) I$.
6. Applying Ampere's Law $\oint \vec{B}\cdot d\vec{l} = \mu_0 I_{\text{enc}}$[cite: 13]:
End Effect: Near the physical open ends of a finite solenoid, field lines diverge outward into half the solid angle, making $B_{\text{end}} = \frac{1}{2}\mu_0 n I$.
Trap 1: The "Zero Line Integral $\implies$ Zero Magnetic Field" Myth
Just like Gauss's Law, $\oint \vec{B}\cdot d\vec{l} = 0$ only guarantees that the net enclosed current is zero ($I_{\text{enc}} = 0$). It does NOT mean $\vec{B} = 0$ everywhere on the loop! An outside current wire produces a non-zero magnetic field on the loop, but $\vec{B}\cdot d\vec{l}$ integrates to positive values along one half and equal negative values along the other half, canceling out to zero[cite: 13].
Trap 2: Turn Density ($n$) vs Total Turns ($N$)
Formula is $B = \mu_0 n I$, where $n = \frac{N}{L}$ is turns per unit length[cite: 13]. If a question states "a solenoid of 500 turns is stretched to double its length keeping current same", $N$ is constant but $L$ is doubled $\implies n' = n/2 \implies B$ is halved! Do not plug total turns $N$ directly into the formula without dividing by length $L$.
Trap 3: Thick Cylindrical Wire: Field Inside ($r < R$)
For a solid wire of radius $R$ carrying uniform current density $J = I/(\pi R^2)$:
• Outside ($r \ge R$): $I_{\text{enc}} = I \implies B = \frac{\mu_0 I}{2\pi r} \propto \frac{1}{r}$[cite: 13].
• Inside ($r < R$): $I_{\text{enc}} = J(\pi r^2) = I\left(\frac{r^2}{R^2}\right) \implies B(2\pi r) = \mu_0 I\frac{r^2}{R^2} \implies \mathbf{B = \frac{\mu_0 I r}{2\pi R^2} \propto r}$!
Field increases linearly from zero at center to maximum at surface, then falls as $1/r$.
Trap 4: Dependence on Solenoid Radius ($R$)
Notice that $B = \mu_0 n I$ contains zero dependence on the cross-sectional radius $R$ of the solenoid[cite: 13]! Whether a solenoid has a radius of $1\text{ cm}$ or $10\text{ cm}$, if turns per metre $n$ and current $I$ are identical, the interior axial magnetic field is exactly the same[cite: 13].
Problem: A solenoid of length $0.5\text{ m}$ has a radius of $1\text{ cm}$ and is made up of $500\text{ turns}$. It carries a current of $5\text{ A}$. What is the magnitude of the magnetic field (i) at the center inside the solenoid, (ii) at one of its extreme ends?
Solution:
1. Calculate turns per unit length ($n$)[cite: 13]:
2. Field inside near the center[cite: 13]:
3. Field at an extreme end:
Problem: A long straight solid copper wire of radius $R = 2.0\text{ mm}$ carries a steady uniformly distributed current of $16\text{ A}$. Find the magnitude of the magnetic field at distance (i) $r_1 = 1.0\text{ mm}$ from the axis of the wire, (ii) $r_2 = 4.0\text{ mm}$ from the axis of the wire.
Solution:
Case (i) Inside the wire ($r_1 = 1.0\text{ mm} < R$):
Current enclosed by Amperian loop of radius $r_1$:
Applying ACL: $B_1(2\pi r_1) = \mu_0 I_{\text{enc}}$:
Case (ii) Outside the wire ($r_2 = 4.0\text{ mm} > R$):
Entire wire current is enclosed ($I_{\text{enc}} = 16\text{ A}$)[cite: 13]:
Symmetry Insight: Notice $B_1 = B_2 = 0.8\text{ mT}$! Because inside $B \propto r$ and outside $B \propto 1/r$, the field at $R/2$ equals the field at $2R$[cite: 13].
Ampere's Law
$\oint \vec{B}\cdot d\vec{l} = \mu_0 I_{\text{enc}}$[cite: 13]
Long Wire Field
$B = \frac{\mu_0 I}{2\pi r}$[cite: 13]
Ideal Solenoid
$B = \mu_0 n I$[cite: 13]
Solenoid End
$B_{\text{end}} = \frac{1}{2}\mu_0 n I$