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(B) The displacement current $I_d$ is given by $I_d = \varepsilon_0 \frac{d\phi_E}{dt}$.Using the Ampere-Maxwell law for a circular path of radius $r$

Vedclass · Accepted Answer

$0.779 	imes 10^{-5} 	ext{ T}$

Vedclass · Answer

(B) The displacement current $I_d$ is given by $I_d = \varepsilon_0 \frac{d\phi_E}{dt}$.Using the Ampere-Maxwell law for a circular path of radius $r$ inside the capacitor: $\oint B \cdot dl = \mu_0 I_d$.Since the electric field is uniform,the flux through a circular area of radius $r$ is $\phi_r = \phi_E \left( \frac{\pi r^2}{A} ight)$,where $A$ is the total area of the plate.Thus,$B(2\pi r) = \mu_0 \varepsilon_0 \frac{d}{dt} \left( \phi_E \frac{\pi r^2}{A} ight) = \mu_0 \varepsilon_0 \frac{r^2}{A} \frac{d\phi_E}{dt} \cdot \pi$.$B = \frac{\mu_0 \varepsilon_0 r}{2A} \frac{d\phi_E}{dt}$.Given $\frac{d\phi_E}{dt} = 7 	imes 10^{14}$,$r = 0.1 	ext{ m}$,$A = 1 	ext{ m}^2$,$\varepsilon_0 = 8.85 	imes 10^{-12}$,$\mu_0 = 4\pi 	imes 10^{-7}$.$B = \frac{(4\pi 	imes 10^{-7})(8.85 	imes 10^{-12})(0.1)}{2(1)} (7 	imes 10^{14}) \approx 7.79 	imes 10^{-6} 	ext{ T} = 0.779 	imes 10^{-5} 	ext{ T}$.

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