First slide

Maxwell's displacement current

Question

A condenser has two conducting plates of radius $10 c m$ separated by a distance of $5 m m$ . It is charged with a constant current of $0.15 A$ . The magnetic field at a point $2 c m$ from the axis in the gap is

Moderate

A

$1.5 \times 10^{- 6} T$
B

$3 \times 10^{- 6} T$
C

$6 \times 10^{- 8} T$
D

$4 \times 10^{- 6} T$

Solution

Consider a loop of radius $' r'$ between the two circular plates, placed axially with them. The area of the loop, $A^{'} = π r^{2}$
By symmetry magnetic field induction $\vec{B}$ is equal in magnitude and is tangential to the circle at every point
In this case, only displacement current $I_{D}$ , will cross the loop. Therefore, using Ampere’s Maxwell law, we have
$\oint \vec{B} . \vec{d l} = μ_{0} I_{D}$
$\Rightarrow 2 π r B = μ_{0} \times (current passing through the area A') \Rightarrow 2 π r B = μ_{0} I \frac{π r^{2}}{π R^{2}} \Rightarrow B = \frac{μ_{0} I_{D}}{2 π r} \times {(\frac{r}{R})}^{2} \Rightarrow B = \frac{μ_{0} I}{2 π} \times \frac{r}{R^{2}} \Rightarrow B = 2 \times 10^{- 7} \times 0.15 \times \frac{2 \times 10^{- 2}}{10^{- 2}} \Rightarrow B = 6 \times 10^{- 8} T$

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