GATE 2024 EE Q.52: solution and answer

GATE 2024 · EEQ.52 · 2 marks · NAT

The given equation represents a magnetic field strength $\overline{H} (r, θ, ϕ)$ in the spherical coordinate system, in free space. Here, $\overset{r}{^}$ and $\hat{θ}$ represent the unit vectors along $r$ and $θ$ , respectively. The value of $P$ in the equation should be ______ (rounded off to the nearest integer).

\overline{H} (r, θ, ϕ) = \frac{1}{r ^{3}} (\overset{r}{^} P cos θ + \hat{θ} sin θ)

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Answer and solution

ANSWER

$P = 2$ . Only this value makes $\nabla \cdot H = 0$ , and it also makes $\nabla \times H = 0$ in the current-free region.

Official answer: 2Source: Official answer key, GATE 2024 EE (IISc Bengaluru)

Every magnetic field obeys Gauss's law for magnetism, $\nabla \cdot B = 0$ . In free space $B = μ_{0} H$ , where $μ_{0} = 4 π \times 1 0^{- 7} H/m$ is a constant, so $\nabla \cdot H = 0$ as well.
Read off the components: $H_{r} = \frac{P cos θ}{r ^{3}}$ , $H_{θ} = \frac{sin θ}{r ^{3}}$ , $H_{ϕ} = 0$ .
Divergence in spherical coordinates: $\nabla \cdot H = \frac{1}{r ^{2}} \frac{\partial}{\partial r} (r^{2} H_{r}) + \frac{1}{r sin θ} \frac{\partial}{\partial θ} (sin θ H_{θ}) + \frac{1}{r sin θ} \frac{\partial H _{ϕ}}{\partial ϕ}$ The last term is zero because $H_{ϕ} = 0$ .
Radial term: $r^{2} H_{r} = \frac{P cos θ}{r}$ , so $\frac{1}{r ^{2}} \frac{\partial}{\partial r} (\frac{P cos θ}{r}) = \frac{1}{r ^{2}} (- \frac{P cos θ}{r ^{2}}) = - \frac{P cos θ}{r ^{4}}$ .
Polar term: $sin θ H_{θ} = \frac{sin ^{2} θ}{r ^{3}}$ , and $\frac{\partial}{\partial θ} (\frac{sin ^{2} θ}{r ^{3}}) = \frac{2 sin θ cos θ}{r ^{3}}$ , so the term is $\frac{1}{r sin θ} \cdot \frac{2 sin θ cos θ}{r ^{3}} = \frac{2 cos θ}{r ^{4}}$ .
The sum must vanish at every $r$ and $θ$ : $\nabla \cdot H = \frac{( 2 - P ) cos θ}{r ^{4}} = 0 \Rightarrow P = 2$
Check: the field region is free space with no current, so $\nabla \times H$ should also be $0$ there. Only the $\hat{ϕ}$ component can be non-zero: $(\nabla \times H)_{ϕ} = \frac{1}{r} [\frac{\partial ( r H _{θ} )}{\partial r} - \frac{\partial H _{r}}{\partial θ}] = \frac{1}{r} [- \frac{2 sin θ}{r ^{3}} + \frac{P sin θ}{r ^{3}}] = \frac{( P - 2 ) sin θ}{r ^{4}}$ This is zero only for $P = 2$ as well. Both conditions agree, and the field is the familiar magnetic-dipole far field, $H \propto \frac{2 cos θ r ^ + sin θ θ ^}{r ^{3}}$ .
Answer: $P = 2$ .

KEY IDEA

A proposed field is a physically valid magnetic field only if its divergence is zero ( $\nabla \cdot B = 0$ , so $\nabla \cdot H = 0$ in a linear homogeneous medium). Setting the divergence to zero fixes the unknown constant. In a current-free region $\nabla \times H = 0$ gives an independent check. The result, $2 cos θ \overset{r}{^} + sin θ \hat{θ}$ over $r^{3}$ , is the magnetic-dipole field.

WATCH OUT FOR

Using the Cartesian-style divergence $\partial H_{r} / \partial r + \partial H_{θ} / \partial θ$ and leaving out the $r^{2}$ and $sin θ$ factors. That gives $- 3 P cos θ / r^{4} + cos θ / r^{3}$ , which cannot vanish for every $r$ and $θ$ , or leads to a wrong value such as $P = 1$ or $P = 3$ . Another slip is dropping the $sin θ$ inside the $θ$ -derivative: differentiating $sin θ$ alone gives $cos θ$ and so $P = 1$ , when the correct derivative of $sin^{2} θ$ gives the factor of 2.

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GATE 2024 EE Q.52: solution and answer

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