A long solenoid of cross-sectional radius R has a thin insulated...

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|\dfrac{dϕ}{dt}\right| = \pi{R}^{2}b = ir + i\left(10r\right)$</p><p>$⇒ i = \dfrac{\pi{R}^{2}b}{11r}$</p><p>The induced electric field in both the parts of the ring is</p><p>${E}_{ind} = \dfrac{R}{2}\dfrac{dB}{dt} = \dfrac{Rb}{2}$</p><p><img alt="" src="https://cdn-question-pool.Quantrex.app/nta_abhyas/jee_main/images/13373_sol_1.JPG" style="width: 130px; height: 163px;" /><br /><br />Let$E$be the field which is developed in each part due to charge accumulation at the junctions, then for the upper half</p><p>$\left({E}_{ind} + E\right)\piR = i\left(10r\right)$</p><p>and for the lower half</p><p>\left({E}_{ind} - E\right)\piR = ir$

$⇒ 2E\piR - 9ir = 0$

$E = \dfrac{9ir}{2\piR}$

$⇒ E = \dfrac{9r}{2\piR} \times \dfrac{\pi{R}^{2}b}{11r} = \dfrac{9}{22}Rb$