Q28 / 68 · NTA Abhyas (JEE Main) 2020 · Physics · Electromagnetic Induction

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
    911Rb
  • Correct answer
    B
    922Rb
  • C
    9 Rb
  • D
    Rb

Solution

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

dϕdt=πR2b=ir+i10r

⇒i=πR2b11r

The induced electric field in both the parts of the ring is

Eind=R2dBdt=Rb2

Figure

Let E be the field which is developed in each part due to charge accumulation at the junctions, then for the upper half

Eind+EπR=i10r

and for the lower half

Eind-EπR=ir

⇒2EπR-9ir=0

E=9ir2πR

⇒E=9r2πR×πR2b11r=922Rb

Hence, the correct answer is option: 'B'.

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