First slide

Maxwell's displacement current

Question

A capacitor made of two circular plates each of radius 12 cm and separated by 5mm. The capacitor is being charged by an external source. The charging current is constant and equal to 0.15A. The magnetic field between the plates at a point 15 cm from the axis is

Moderate

A

$2 \times 10^{- 2} T$
B

$1 \times 10^{- 2} T$
C

$1 \times 10^{- 7} T$
D

$2 \times 10^{- 7} T$

Solution

Consider a loop of radius ’r’ between the two circular plates placed coaxially with them . Area of the loop $A' = π r^{2}$
In this case, only displacement current $I_{D}$ will cross the loop. Therefore using Ampere’s Maxwell law, we have
$\oint \vec{B} . \vec{d l} = μ_{o} I_{D}$
$\Rightarrow 2 π r B = μ_{o} x (c u r r e n t p a s \sin g t h r o u g h t h e a r e a A')$
$\Rightarrow 2 π r B = μ_{o} I \times \frac{π r^{2}}{π R^{2}} [f o r r < R]$
$\Rightarrow B = \frac{μ_{o} I}{2 π r} \times \frac{r^{2}}{R^{2}}$
But for $r > R$ , $B = \frac{μ_{0} I}{2 π r}$
$∴ B = \frac{4 π \times 10^{- 7} \times 0.15}{2 π \times 0.15} = 2 \times 10^{- 7} T$

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