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NEET-XII-Physics

38: Electromagnetic Induction

with Solutions - page 5
Qstn# iv-7-a Prvs-QstnNext-Qstn
  • #7-a
    its removal, (b) its restoration and (c) its motion.
    Ans : During the removal the emf induced in the coil,
    e = 50 V
    time taken, t = 0.25 s
    current in the coil, `` i=\frac{e}{\,\mathrm{\,R\,}}=2\,\mathrm{\,A\,}``
    Thus, the thermal energy developed is given by
    H = I2RT
    = 4 × 25 × 0.25 = 25 J (b) During the restoration of the coil,
    emf induced in it, e = 50 V
    time taken, t = 0.25 s
    current in the coil, `` i=\frac{e}{\,\mathrm{\,R\,}}=2\,\mathrm{\,A\,}``
    Thus, the thermal energy developed is given by
    H = i2RT = 25 J (c) We know that energy is a scalar quantity. Also, the net thermal energy is the algebraic sum of the two energies calculated.
    ∴ Net thermal energy developed
    = 25 J + 25 J = 50 J
    Page No 306:
  • #7-b
    its restoration and
    Ans : During the restoration of the coil,
    emf induced in it, e = 50 V
    time taken, t = 0.25 s
    current in the coil, `` i=\frac{e}{\,\mathrm{\,R\,}}=2\,\mathrm{\,A\,}``
    Thus, the thermal energy developed is given by
    H = i2RT = 25 J
  • #7-c
    its motion.
    Ans : We know that energy is a scalar quantity. Also, the net thermal energy is the algebraic sum of the two energies calculated.
    ∴ Net thermal energy developed
    = 25 J + 25 J = 50 J
    Page No 306: