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

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 = b t , where b is a constant. The magnitude of the electric field strength in the ring (other than the induced electric field) is

TestHub · Accepted Answer

Let us assume that the current in the ring is $i$ and the resistance of the two halves are $r$ and $10 r$ , then by faraday's law

$|\frac{d ϕ}{d t}| = π R^{2} b = i r + i (10 r)$

$\Rightarrow i = \frac{π R^{2} b}{11 r}$

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

$(E_{i n d} + E) π R = i (10 r)$

and for the lower half

$(E_{i n d} - E) π R = i r$

$\Rightarrow 2 E π R - 9 i r = 0$

$E = \frac{9 i r}{2 π R}$

$\Rightarrow E = \frac{9 r}{2 π R} \times \frac{π R^{2} b}{11 r} = \frac{9}{22} R b$

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