Question

A long solenoid of cross-sectional radius $R$ has a thin insulated wire ring of uniform cross-section tightly put on its winding. The ring is made out of two different material such that one half of the ring has a resistance $10$ times that of the other half. The magnetic induction produced by the solenoid varies with time as $B=bt$ , where $b$ is a constant. The magnitude of the electric field strength in the ring (other than the induced electric field) is

• A. $\frac{9}{1 1}Rb$
• B. $\frac{9}{2 2}Rb$
• C. $9 \, Rb$
• D. $Rb$

Answer 1

Answer: (B)

Let us assume that the current in the ring is $i$ and the resistance of the two halves are $r$ and $10r$ , then by faraday's law
$\left|\frac{d \phi}{d t}\right|=\pi R^{2}b=ir+i\left(10 r\right)$
$\Rightarrow i=\frac{\pi R^{2} b}{11 r}$
The induced electric field in both the parts of the ring is
$E_{ind}=\frac{R}{2}\frac{d B}{d t}=\frac{R b}{2}$

Let $E$ be the field which is developed in each part due to charge accumulation at the junctions, then for the upper half
$\left(E_{i n d} + E\right)\pi R=i\left(10 r\right)$
and for the lower half
$\left(E_{i n d} - E\right)\pi R=ir$
$\Rightarrow 2E\pi R-9ir=0$
$E=\frac{9 i r}{2 \pi R}$
$\Rightarrow E=\frac{9 r}{2 \pi R}\times \frac{\pi R^{2} b}{1 1 r}=\frac{9}{2 2}Rb$