Question

The electric flux $\left({\varphi }_E\right)$ in a space varies with time $\left(t\right)$ as, ${\varphi }_E=250t^2+100t-120$, where $t$ is in $\text{second}$. The magnitude of displacement current at $t=10\text{s}$ is -

Take ${ϵ}_0=8.85\times {10}^{-12}{\text{C}}^2/{\text{N-m}}^2$

• A. $0.01\mu \text{A}$
• B. $0.045\mu \text{A}$
• C. $0.1\mu \text{A}$
• D. $45\mu \text{A}$

Answer 1

The correct option is B $0.045\mu \text{A}$

The displacement current is given by,

$I_d={ϵ}_o\frac{d{\varphi }_E}{dt}$

Here,

${\varphi }_E=250t^2+100t-120$

So,

$\frac{d{\varphi }_E}{dt}=500t+100$

At $t=10\text{s}$ ,

$\frac{d{\varphi }_E}{dt}=500\times 10+100=5100\text{V-m}$

Now,

$\therefore I_d=5100\times 8.85\times {10}^{-12}\approx 4.5\times {10}^{-8}\text{A}=0.045\mu \text{A}$

Hence, option $\left(B\right)$ is the correct answer.