Question

An electric field of $300V/m$ is confined to a circular area $10cm$ in diameter. If the field is increasing at the rate of $20V/m/s$, the magnitude field at a point $15cm$ from the centre of the circle will be:

• A. $1.85\times {10}^{-15}T$
• B. $1.85\times {10}^{-16}T$
• C. $1.85\times {10}^{-17}T$
• D. $5.55\times {10}^{-18}T$

Answer 1

The correct option is D $5.55\times {10}^{-18}T$

Area of electric field $A=\pi (\frac{d}{2}{)}^2$

Change in electric flux = change in Electric field $\times$ area

Change in electric flux per unit time $\frac{d\varphi }{dt}=20\times 3.14$$\times 5$$\times 5$$\times {10}^{-4}$

From ampere's maxwell law,

$\oint \overrightarrow{B}.\overrightarrow{dl}={\mu }_o{\epsilon }_o\frac{d\varphi }{dt}$

$B=\frac{{\mu }_o{ϵ}_o\frac{d\varphi }{dt}}{2\pi r}=\frac{20\times \pi \times 25\times {10}^{-4}}{(3\times {10}^8{)}^2\times 2\times \pi \times 0.05}=5.55\times {10}^{-18}T$