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GATE 2016 EC – Question 61

Electromagnetics · Maxwell's Equations · 2 marks · Multiple choice

The current density in a medium is given by

$$\vec{J} = \frac{400\sin\theta}{2\pi(r^2 + 4)}\hat{a}_r \text{ Am}^{-2}$$

The total current and the average current density flowing through the portion of a spherical surface $r = 0.8$ m, $\frac{\pi}{12} \leq \theta \leq \frac{\pi}{4}$, $0 \leq \phi \leq 2\pi$ are given, respectively, by

  1. 15.09 A, 12.86 Am$^{-2}$
  2. 18.73 A, 13.65 Am$^{-2}$
  3. 12.86 A, 9.23 Am$^{-2}$
  4. 10.28 A, 7.56 Am$^{-2}$

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Correct answer: (D) 10.28 A, 7.56 Am$^{-2}$

Explanation

The surface element on a sphere is $r^2\sin\theta\,d\theta\,d\phi$, so $I = \int\int \frac{400\sin\theta}{2\pi(r^2 + 4)} r^2\sin\theta\,d\theta\,d\phi = \frac{400 r^2}{r^2 + 4}\int_{\pi/12}^{\pi/4}\sin^2\theta\,d\theta$. With $r = 0.8$ this is $55.2 \times 0.1368 = 7.55$ A. The area of the surface portion is $2\pi r^2(\cos\frac{\pi}{12} - \cos\frac{\pi}{4}) = 1.04$ m$^2$, so the average density is $\frac{7.55}{1.04} = 7.25$ A/m$^2$. No option gives these values, and the official key awards marks to all candidates.