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NEET-XII-Physics

38: Electromagnetic Induction

with Solutions - page 10
Qstn# iv-42-a Prvs-QstnNext-Qstn
  • #42-a
    both the wires move towards right and (b) if P1Q1 moves towards left but P2Q2 moves towards right.
    Figure
    Ans : When both wires move in same direction:
    The sliding wires constitute two parallel sources of emf.
    The net emf is given by
    e = Blv
    ⇒ e = (1 × 4 × 10-2 ) × 5 × (10-2)
    = 20 × 10-4 V
    The resistance of the sliding wires is 2 Ω.
    ∴ Net resistance = `` \frac{2\times 2}{2+2}`` + 19 = 20 Ω
    Net current through 19 Ω = `` \frac{2\times {10}^{-4}}{20}`` = 0.1 mA (b) When both wires move in opposite directions with the same speed, the direction of the emf induced in both of them is opposite. Thus, the net emf is zero.
    ∴ Net current through 19 Ω = 0
    Page No 309:
  • #42-b
    if P1Q1 moves towards left but P2Q2 moves towards right.
    Figure
    Ans : When both wires move in opposite directions with the same speed, the direction of the emf induced in both of them is opposite. Thus, the net emf is zero.
    ∴ Net current through 19 Ω = 0
    Page No 309: