Question

Consider a medium of uniform resistivity with a pair of source and sink electrodes separated by a distance \( L \), as shown in the figure. The fraction of the input current \( (I) \) that flows horizontally \( (I_x) \) across the median plane between depths \( z_1 = \frac{L}{2} \) and \( z_2 = \frac{L\sqrt{3}}{2} \), is given by \( \frac{I_x}{I} = \frac{L}{\pi} \int_{z_1}^{z_2} \frac{dz}{(L^2/4 + z^2)} \). The value of \( \frac{I_x}{I} \) is equal to 

 

Hint:

Remember the integral of \( \frac{1}{a^2 + x^2} \) and the values of \( \arctan \) for standard angles. Pay close attention to the limits of integration.

Answer 1

Correct Answer: 0.16

Step 1: Substitute the limits of integration. \[ \frac{I_x}{I} = \frac{L}{\pi} \int_{L/2}^{L\sqrt{3}/2} \frac{dz}{(L^2/4 + z^2)} \] Step 2: Evaluate the integral.
Using the integral formula \( \int \frac{dx}{a^2 + x^2} = \frac{1}{a} \arctan(\frac{x}{a}) + C \) with \( a = L/2 \): \[ \int \frac{dz}{(L^2/4 + z^2)} = \frac{1}{L/2} \arctan\left(\frac{z}{L/2}\right) = \frac{2}{L} \arctan\left(\frac{2z}{L}\right) \] Step 3: Apply the limits of integration. \[ \frac{I_x}{I} = \frac{L}{\pi} \left[ \frac{2}{L} \arctan\left(\frac{2z}{L}\right) \right]_{L/2}^{L\sqrt{3}/2} = \frac{2}{\pi} \left[ \arctan\left(\frac{2(L\sqrt{3}/2)}{L}\right) - \arctan\left(\frac{2(L/2)}{L}\right) \right] \] \[ \frac{I_x}{I} = \frac{2}{\pi} \left[ \arctan(\sqrt{3}) - \arctan(1) \right] \] Step 4: Evaluate the arctangent values. \[ \arctan(\sqrt{3}) = \frac{\pi}{3}, \quad \arctan(1) = \frac{\pi}{4} \] Step 5: Calculate the final value. \[ \frac{I_x}{I} = \frac{2}{\pi} \left[ \frac{\pi}{3} - \frac{\pi}{4} \right] = \frac{2}{\pi} \left[ \frac{4\pi - 3\pi}{12} \right] = \frac{2}{\pi} \left[ \frac{\pi}{12} \right] = \frac{1}{6} \] Step 6: Convert to decimal and round off to two decimal places. \[ \frac{1}{6} \approx 0.1666... \approx 0.17 \]