Question

A surface current density \(\mathbf{K} = ae^{-y}\) exists on a thin strip of width b, as shown in the figure below. The associated surface current is
(a is a constant of appropriate dimensions)

• A. \(a(1 - e^{-b})\)
• B. \(a(1 + e^{-b})\)
• C. \(a(e^{-b} - 1)\)
• D. \(a(e^b + e^{-b})\)

Hint:

Remember the relationship between different types of currents and densities. - Line current \(I\) (Amperes). - Surface current density \(K\) (Amperes/meter). \(I = \int K \, dL_{\perp}\). - Volume current density \(J\) (Amperes/meter\(^2\)). \(I = \int J \, dA_{\perp}\). Correctly identifying which quantity is given and how to integrate it is key.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
The problem asks for the total surface current (\(I\)) flowing through a strip, given the surface current density (\(\mathbf{K}\)). Surface current density \(\mathbf{K}\) is defined as the current per unit length perpendicular to the flow. To find the total current, we need to integrate the current density over the width of the strip.
Step 2: Key Formula or Approach:
The total current \(I\) flowing through a surface is found by integrating the surface current density \(\mathbf{K}\) over the path perpendicular to the current flow. If the current flows in the x-direction and varies with y, the infinitesimal current \(dI\) through a small width \(dy\) is \(dI = K(y) \, dy\). The total current is: \[ I = \int K(y) \, dy \] The integration is performed over the width of the strip.
Step 3: Detailed Explanation:
1. Identify the given quantities: The surface current density is given as \(\mathbf{K} = ae^{-y}\). The diagram indicates that the current is flowing in the positive x-direction (\(\mathbf{K} = ae^{-y} \hat{i}\)). The current density is uniform in x but varies with y. The strip extends from \(y=0\) to \(y=b\). 2. Set up the integral for the total current: To find the total current \(I\) passing through the strip, we need to integrate the magnitude of the current density, \(K(y) = ae^{-y}\), over the width of the strip, which is along the y-axis from 0 to b. \[ I = \int_{0}^{b} K(y) \, dy \] \[ I = \int_{0}^{b} ae^{-y} \, dy \] 3. Evaluate the integral: The constant \(a\) can be taken out of the integral. \[ I = a \int_{0}^{b} e^{-y} \, dy \] The integral of \(e^{-y}\) is \(-e^{-y}\). \[ I = a \left[ -e^{-y} \right]_{0}^{b} \] Now, apply the limits of integration: \[ I = a \left( (-e^{-b}) - (-e^{-0}) \right) \] Since \(e^0 = 1\): \[ I = a (-e^{-b} + 1) \] \[ I = a(1 - e^{-b}) \] Step 4: Final Answer:
The associated surface current is \(a(1 - e^{-b})\). This corresponds to option (A).