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NEET-XII-Physics

38: Electromagnetic Induction

with Solutions - page 10
Qstn# iv-44 Prvs-QstnNext-Qstn
  • #44
    Consider the situation shown in figure. The wire PQ has a negligible resistance and is made to slide on the three rails with a constant speed of 5 cm s-1. Find the current in the 10 Ω resistor when the switch S is thrown to (a) the middle rail (b) the bottom rail.
    Figure
    Ans : Given:
    Magnetic field, B = 1 T
    Velocity of the sliding wire, v = 5 × 10-2 m/s
    Resistance of the connected resistor, R = 10 Ω (a) When the switch is thrown to the middle rail:
    Length of the sliding wire = 2 × 10-2 m
    Induced emf, E = Bvl
    = 1 × (5 × 10-2) × (2 × 10-2) V
    = 10 × 10-4 = 10-3 V
    Current in the 10 Ω resistor is given by
    `` i=\frac{E}{R}``
    `` =\frac{{10}^{-3}}{10}={10}^{-4}=0.1\,\mathrm{\,mA\,}`` (b) When the switch is thrown to the bottom rail:
    The length of the sliding wire becomes 4 × 10-2 m.
    The induced emf is given by
    E = Bvl'
    = 1 × (5 × 10-2) × (4 × 10-2)
    = 20 × 10-4 V
    Now,
    Current, i = `` \frac{20\times {10}^{-4}}{10}`` A
    = 2 × 10-4 A = 0.2 mA
    Page No 309:
  • #44-a
    the middle rail
    Ans : When the switch is thrown to the middle rail:
    Length of the sliding wire = 2 × 10-2 m
    Induced emf, E = Bvl
    = 1 × (5 × 10-2) × (2 × 10-2) V
    = 10 × 10-4 = 10-3 V
    Current in the 10 Ω resistor is given by
    `` i=\frac{E}{R}``
    `` =\frac{{10}^{-3}}{10}={10}^{-4}=0.1\,\mathrm{\,mA\,}``
  • #44-b
    the bottom rail.
    Figure
    Ans : When the switch is thrown to the bottom rail:
    The length of the sliding wire becomes 4 × 10-2 m.
    The induced emf is given by
    E = Bvl'
    = 1 × (5 × 10-2) × (4 × 10-2)
    = 20 × 10-4 V
    Now,
    Current, i = `` \frac{20\times {10}^{-4}}{10}`` A
    = 2 × 10-4 A = 0.2 mA
    Page No 309: