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NEET-XII-Physics

38: Electromagnetic Induction

with Solutions - page 10
Qstn# iv-43 Prvs-QstnNext-Qstn
  • #43
    Suppose the 19 Ω resistor of the previous problem is disconnected. Find the current through P2Q2 in the two situations (a) and (b) of that problem.
    Ans : (a) When the wires move in the same direction, their polarity remains the same. The circuit remains incomplete. Therefore, no current flows in the circuit. (b) When the wires move in opposite directions, their polarities are reversed. Thus, current flows in the circuit.
    `` {V}_{{\,\mathrm{\,P\,}}_{2}{\,\mathrm{\,Q\,}}_{2}}=Blv``
    = 1 × 0.04 × 0.05
    = 2 × 10-3 V
    R = 2 Ω
    Current in the circuit is given by
    `` i=\frac{2\times {10}^{-3}}{2}``
    = 1 × 10-3 A = 1 mA
    Page No 309:
  • #43-a
    and
    Ans : When the wires move in the same direction, their polarity remains the same. The circuit remains incomplete. Therefore, no current flows in the circuit.
  • #43-b
    of that problem.
    Ans : When the wires move in opposite directions, their polarities are reversed. Thus, current flows in the circuit.
    `` {V}_{{\,\mathrm{\,P\,}}_{2}{\,\mathrm{\,Q\,}}_{2}}=Blv``
    = 1 × 0.04 × 0.05
    = 2 × 10-3 V
    R = 2 Ω
    Current in the circuit is given by
    `` i=\frac{2\times {10}^{-3}}{2}``
    = 1 × 10-3 A = 1 mA
    Page No 309: