GSO001 Problem 20

This problem applies Ampere's law (in the integral form via Stokes' theorem) from Maxwell's equations to find the current from the magnetic field distribution — a vector calculus application.

The most straightforward approach is to apply the integral form of Ampere's law. Let the closed curve $C$ bounding the disk $S$ be the circle of radius $R = \frac{1}{\sqrt{\alpha}}$ (i.e., $x^2 + y^2 = R^2$ , $z = 0$ ). The line integral of the magnetic field along $C$ is related to the total current $I$ through $S$ by:

\oint_C \mathbf{B} \cdot d\mathbf{l} = \mu_0 I

On the circle $C$ , since $x^2 + y^2 = R^2 = \frac{1}{\alpha}$ , the exponent in $\mathbf{B}$ is constant:

e^{-\alpha(x^2 + y^2)} = e^{-\alpha(1/\alpha)} = e^{-1}

Therefore, the magnetic field along the path is:

\mathbf{B} = B_0 e^{-1} (-y \mathbf{i} + x \mathbf{j})

Using polar coordinates $(r, \theta)$ , the infinitesimal displacement vector $d\mathbf{l}$ on the circle is $dx = -R\sin\theta \, d\theta = -y \, d\theta$ , $dy = R\cos\theta \, d\theta = x \, d\theta$ , so:

d\mathbf{l} = dx \mathbf{i} + dy \mathbf{j} = (-y \mathbf{i} + x \mathbf{j}) d\theta

The dot product $\mathbf{B} \cdot d\mathbf{l}$ is:

\begin{aligned} \mathbf{B} \cdot d\mathbf{l} &= B_0 e^{-1} (-y \mathbf{i} + x \mathbf{j}) \cdot (-y \mathbf{i} + x \mathbf{j}) d\theta \\ &= B_0 e^{-1} (y^2 + x^2) d\theta \\ &= B_0 e^{-1} R^2 d\theta \end{aligned}

Integrating from $\theta = 0$ to $2\pi$ :

\oint_C \mathbf{B} \cdot d\mathbf{l} = \int_0^{2\pi} B_0 e^{-1} R^2 d\theta = 2\pi R^2 B_0 e^{-1} = \frac{2\pi B_0}{\alpha e}

By Ampere's law, the total current $I$ is:

I = \frac{1}{\mu_0} \oint_C \mathbf{B} \cdot d\mathbf{l} = \frac{2\pi B_0}{\mu_0 \alpha e}

Substituting the given numerical values:

\begin{aligned} I &= \frac{2\pi \times (1.37 \times 10^{-2})}{(4\pi \times 10^{-7}) \times (2.50 \times 10^3) \times e} \\ &= \frac{2\pi \times 1.37 \times 10^{-2}}{10000\pi \times 10^{-7} \times e} \\ &= \frac{2.74 \pi \times 10^{-2}}{10^{-3} \pi \times e} \\ &= \frac{27.4}{e} \\ &= \frac{274}{10 e} \\ &= \frac{137}{5 e} \text{ A} \end{aligned}

From the problem's input format, $I = \frac{X}{e}$ gives $X = \frac{137}{5}$ .

This is already an irreducible fraction, so $A = 137$ , $B = 5$ .

A + B = 137 + 5 = 142

Answer: 142

GSO001 Problem 20

Current Through a Cylindrical Region in a Static Magnetic Field

Problem Statement

Constraints

Input Format

Solution