Ampère's Law: Wedge Path Line Integral for Wire Current

The image displays a cross-section of a cylindrical wire with radius $a$ carrying a total current $I$ directed out of the page. Overlaid on this is an Amperian path in the shape of a circular wedge (sector). The wedge has two radial sides of length $R$ and a circular arc subtended by an angle $\theta$. Crucially, the diagram shows that $R > a$, meaning the wedge-shaped path extends beyond the boundary of the wire.

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Answer

The correct option is (A). By Ampère’s Law, the line integral of the magnetic field around a closed path is proportional to the total current enclosed by that path, which in this case is a fraction of the total current determined by the ratio of the wedge angle $\theta$ to the full $2\pi$ radians.

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Explanation

1. Knowns and Find

   • Known: Total current $I$ (uniform), wire radius $a$, path angle $\theta$, path radius $R$, and the condition $R > a$.
   • Find: The value of the circulation integral $\oint \mathbf{B} \cdot d\mathbf{l}$.

2. Ampère’s Law Statement According to Ampère’s Law, the line integral of the magnetic field $\mathbf{B}$ around any closed loop is equal to $\mu_0$ times the total current $I_{\text{enc}}$ passing through the surface bounded by the loop: $\oint \mathbf{B} \cdot d\mathbf{l} = \mu_0 I_{\text{enc}}$ The line integral is exactly equal to the vacuum permeability times the enclosed current.

3. Determining Enclosed Current ($I_{\text{enc}}$) The current $I$ is uniformly distributed over the cross-section of the wire (a circle of radius $a$). Since the path radius $R$ is greater than the wire radius $a$, the path encloses a "slice" of the entire wire. The area of the entire wire is $A_{\text{total}} = \pi a^2$. The area of the part of the wire enclosed by the wedge is a sector of radius $a$ and angle $\theta$: $A_{\text{enc}} = \frac{\theta}{2\pi} (\pi a^2) = \frac{\theta a^2}{2}$ The ratio of the enclosed area to the total area determines the enclosed current: $\frac{I_{\text{enc}}}{I} = \frac{A_{\text{enc}}}{A_{\text{total}}} = \frac{\frac{\theta}{2\pi} \pi a^2}{\pi a^2} = \frac{\theta}{2\pi}$ The enclosed current is the fraction of the total circle represented by the angle theta.

4. Substitution and Result Substitute the expression for $I_{\text{enc}}$ back into Ampère’s Law: $\oint \mathbf{B} \cdot d\mathbf{l} = \mu_0 \left( \frac{\theta}{2\pi} I \right) = \frac{\mu_0 \theta I}{2\pi}$ This shows the circulation depends only on the angle and the total current, not the radii $a$ or $R$ (as long as $R \geq a$).

5. Dimensional Analysis (Unit Check) The units of $\mu_0$ are $T \cdot m / A$. Multiplying by current $A$ gives $T \cdot m$. The angle $\theta$ is dimensionless (radians). Thus, the result has units of Tesla-meters, which is correct for a line integral of a magnetic field.

Option	Formula	Reason for Evaluation
(A)	$\frac{\mu_0 \theta I}{2\pi}$	Correct. Properly applies the ratio of the sector to the full circle.
(B)	$\frac{\mu_0 \theta I}{2\pi^2 a^2}$	Incorrect. Dimensions are wrong; includes $1/L^2$.
(C)	$\frac{\mu_0 \theta I}{2\pi^2 R^2}$	Incorrect. Dependence on $R$ is wrong for an enclosed current problem.
(D)/(E)	$\frac{\mu_0 I}{\pi a^2} \dots$	Incorrect. These represent current density proportions, not the line integral.

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Final Answer

The value of the line integral is determined by the portion of the total current enclosed within the angular width of the path. $\boxed{\frac{\mu_0 \theta I}{2\pi}}$

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Common Mistakes

• Including $R$ in the final result: Students often think the size of the path ($R$) affects the integral. However, per Ampère's Law, once the path is outside the current-carrying region ($R > a$), the integral only depends on the total current enclosed, not the distance.
• Confusing Current Density with the Integral: Some might attempt to find $B$ at distance $R$ first and then integrate. While possible, it is much slower than recognizing that $\oint \mathbf{B} \cdot d\mathbf{l}$ is, by definition, $\mu_0 I_{\text{enc}}$.

Related Topics: Ampère's Law, Magnetic Flux, Current Density ($J = I/A$), Gauss's Law for Magnetism.