Question

Consider the cylindrical region of the magnetic field shown in the figure. Region $I$ and $II$ have fields directed perpendicularly outward and inward, respectively. Fields are varying with time as

Region $I:B_1=3B_{0}t$

Region $II:B_2=B_{0}t$

Find the ratio $\frac{r_1}{r_2}$ providing there is no net induced electric field in the region outside the magnetic field.

• A. $0.5$
• B. $0.6$
• C. $0.3$
• D. $0.2$

Answer 1

The correct option is A $0.5$

The induced electric field in a closed loop placed in a varying magnetic field is given by,

$\oint \overrightarrow{E}\cdot \overrightarrow{dl}=-\frac{d{\varphi }_B}{dt}$

Net magnetic flux,

${\varphi }_B=B_1\pi r_1^2-B_2\pi (r_2^2-r_1^2)$

$=3B_{0}t\cdot \pi r_1^2-B_{0}t\pi r_2^2+B_{0}t\pi r_1^2$

$=4B_{0}t\pi r_1^2-B_{0}t\pi r_2^2$

$=B_{0}t\pi (4r_1^2-r_2^2)$

$\because$ Electric field is zero in the region outside the magnetic field,

$\therefore \frac{d{\varphi }_B}{dt}=0$

$\Rightarrow B_0\pi (4r_1^2-r_2^2)=0$

$\Rightarrow 4r_1^2-r_2^2=0$

$\Rightarrow \frac{r_1}{r_2}=\frac{1}{2}=0.5$

Hence, option $\left(A\right)$ is correct.

Note: For a region outside the magnetic field i.e. $r>R$ $\oint \overrightarrow{E}\cdot \overrightarrow{dl}=-\frac{d\varphi }{dt}$ $\Rightarrow E\times 2\pi r=-\frac{d\varphi }{dt}$ $\Rightarrow E=-\frac{d\varphi }{dt}.\frac{1}{2\pi r}$ Therefore, electric field even exists (due to the change in flux inside the region) in the region where magnetic field is not present.