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

Step-by-step solution

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 d ϕ d t = π R 2 b = i r + i 10 r ⇒ i = π R 2 b 11 r The induced electric field in both the parts of the ring is E ind = R 2 d B d t = 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 ⇒ 2 E π R - 9 i r = 0 E = 9 i r 2 π R ⇒ E = 9 r 2 π R × π R 2 b 1