GATE 2022 EE Q.23: solution and answer

GATE 2022 · EEQ.23 · 1 mark · MCQ

A long conducting cylinder having a radius 'b' is placed along the z axis. The current density is $J = J_{a} r^{3} \overset{z}{^}$ for the region $r < b$ where $r$ is the distance in the radial direction. The magnetic field intensity ( $H$ ) for the region inside the conductor (i.e. for $r < b$ ) is

(A)
$\frac{J _{a}}{4} r^{4}$
(B)
$\frac{J _{a}}{3} r^{3}$
(C)
$\frac{J _{a}}{5} r^{4}$
(D)
$J_{a} r^{3}$

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Answer and solution

ANSWER

For $r < b$ , $H = \frac{J _{a}}{5} r^{4} \overset{a}{^}_{ϕ}$ . The answer is option (C).

Official answer: CSource: Official answer key, GATE 2022 EE (IIT Kharagpur)

Symmetry. The current flows along $\overset{z}{^}$ and its density depends only on $r$ . So $H$ is purely azimuthal, $H = H_{ϕ} (r) \overset{a}{^}_{ϕ}$ , and has the same magnitude at every point of a circle of radius $r$ centred on the axis.
Ampère's circuital law on a circle of radius $r < b$ : $\oint H \cdot d l = I_{enc}$ . Because $H_{ϕ}$ is constant on the circle and parallel to $d l$ , this gives $H_{ϕ} (2 π r) = I_{enc} (r)$ .
Enclosed current. The current density is not uniform, so integrate it over the disc of radius $r$ . The area element of a thin ring is $d S = 2 π r^{'} d r^{'}$ : $I_{enc} = \int_{0}^{r} J_{a} r^{'3} (2 π r^{'}) d r^{'} = 2 π J_{a} \int_{0}^{r} r^{'4} d r^{'} = \frac{2 π J _{a} r ^{5}}{5}$
Solve for $H_{ϕ}$ : $H_{ϕ} = \frac{I _{enc}}{2 π r} = \frac{2 π J _{a} r ^{5} /5}{2 π r} = \frac{J _{a}}{5} r^{4}$
Check with the point form $\nabla \times H = J$ . In cylindrical coordinates the $z$ -component is $\frac{1}{r} \frac{d}{d r} (r H_{ϕ}) = \frac{1}{r} \frac{d}{d r} (\frac{J _{a} r ^{5}}{5}) = J_{a} r^{3}$ , which is the given $J_{z}$ .
So inside the conductor $H = \frac{J _{a}}{5} r^{4} \overset{a}{^}_{ϕ}$ . That is option (C).

KEY IDEA

When the current density varies with $r$ , find the enclosed current by integrating over thin rings of area $2 π r^{'} d r^{'}$ . Then apply Ampère's law on a circle of radius $r$ : $H_{ϕ} (2 π r) = I_{enc}$ . A current density that goes as $r^{n}$ gives $H_{ϕ} = \frac{J _{a} r ^{n + 1}}{n + 2}$ , so here ( $n = 3$ ) the result is $\frac{J _{a} r ^{4}}{5}$ .

WATCH OUT FOR

The first mistake is treating $J$ as uniform and writing $I_{enc} = J π r^{2}$ . That gives $\frac{J _{a} r ^{4}}{2}$ . The second is leaving out the factor $r^{'}$ in the ring area $d S = 2 π r^{'} d r^{'}$ , so that $\int r^{'3} d r^{'}$ is integrated instead of $\int r^{'4} d r^{'}$ . That gives $\frac{J _{a} r ^{3}}{4}$ . Neither result is an option, and option (A) pairs the power $r^{4}$ with the wrong factor $\frac{1}{4}$ . Option (D) is just the magnitude of $J$ , which mixes up the current density with the field it produces.

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GATE 2022 EE Q.23: solution and answer

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