Question

For the situation described in the figure, the magnetic field changes with time according to, $B=(2.00t^3-4.00t^2+0.8)\text{T}$. What is the magnitude of electric field at point ${\text{P}}_1$ when $t=0.02\text{s}$ and $r_1=0.02\text{m}$.

• A. $1\text{N/C}$
• B. $2.5\text{N/C}$
• C. $3\text{N/C}$
• D. $0.3\text{N/C}$

Answer 1

The correct option is D $0.3\text{N/C}$

We know that the induced electric field at point ${\text{P}}_1$ is given by,

$E=\frac{r_1}{2}\frac{dB}{dt}$

Substituting the value of $\left(B\right)$,

$E=\frac{r_1}{2}\frac{d}{dt}[2t^3-4t^2+0.8]$

$E=\frac{r_1}{2}\times (6t^2-8t)$

Here, $r_1=0.02\text{m}$ and $t=3\text{s}$

$\therefore E=\frac{0.02}{2}\times \left[\right(6\times 3^2)-(8\times 3\left)\right]=0.3\text{N/C}$

Hence, $\left(\text{D}\right)$ is the correct answer.