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NEET-XII-Physics

38: Electromagnetic Induction

with Solutions - page 7
Qstn# iv-22 Prvs-QstnNext-Qstn
  • #22
    A closed coil having 100 turns is rotated in a uniform magnetic field B = 4.0 × 10-4 T about a diameter which is perpendicular to the field. The angular velocity of rotation is 300 revolutions per minute. The area of the coil is 25 cm2 and its resistance is 4.0 Ω. Find (a) the average emf developed in half a turn from a position where the coil is perpendicular to the magnetic field, (b) the average emf in a full turn and (c) the net charge displaced in part
    (a).
    Ans : Given:
    Number of turns in the coil, n = 100 turns
    Magnetic field, B = 4 × 10-4
    Area of the loop, A = 25 cm2 = 25 × 10-4 m2 (a) When the coil is perpendicular to the field:
    ϕ1 = nBA
    When the coil goes through the half turn:
    ϕ2 = nBA cos 180° = -nBA
    ∴ Δϕ = 2nBA
    When the coil undergoes 300 revolutions in 1 minute, the angle swept by the coil is
    300 × 2π rad/min = 10π rad/s
    10π rad is swept in 1 s.
    π rad is swept in `` \left(\frac{1}{10\,\mathrm{\,\pi \,}}\right)\,\mathrm{\,\pi \,}=\frac{1}{10}\,\mathrm{\,s\,}``
    `` e=\frac{d\,\mathrm{\,\varphi \,}}{dt}=\frac{2nBA}{dt}``
    `` =\frac{2\times 100\times 4\times {10}^{-4}\times 25\times {10}^{-4}}{1/10}``
    `` =2\times {10}^{-3}\,\mathrm{\,V\,}`` (b) ϕ1 = nBA, ϕ2 = nBA (θ = 360°)
    Δϕ = 0, thus emf induced will be zero. (c) The current flowing in the coil is given by
    `` i=\frac{e}{R}=\frac{2\times {10}^{-3}}{4}=\frac{1}{2}\times {10}^{-3}``
    = 0.5 × 10-3 = 5 × 10-4 A
    Hence, the net charge is given by
    Q = idt = 5 × 10-4 × `` \frac{1}{10}``
    = 5 × 10-5 C
    Page No 307:
  • #22-a
    the average emf developed in half a turn from a position where the coil is perpendicular to the magnetic field,
    Ans : When the coil is perpendicular to the field:
    ϕ1 = nBA
    When the coil goes through the half turn:
    ϕ2 = nBA cos 180° = -nBA
    ∴ Δϕ = 2nBA
    When the coil undergoes 300 revolutions in 1 minute, the angle swept by the coil is
    300 × 2π rad/min = 10π rad/s
    10π rad is swept in 1 s.
    π rad is swept in `` \left(\frac{1}{10\,\mathrm{\,\pi \,}}\right)\,\mathrm{\,\pi \,}=\frac{1}{10}\,\mathrm{\,s\,}``
    `` e=\frac{d\,\mathrm{\,\varphi \,}}{dt}=\frac{2nBA}{dt}``
    `` =\frac{2\times 100\times 4\times {10}^{-4}\times 25\times {10}^{-4}}{1/10}``
    `` =2\times {10}^{-3}\,\mathrm{\,V\,}``
  • #22-b
    the average emf in a full turn and
    Ans : ϕ1 = nBA, ϕ2 = nBA (θ = 360°)
    Δϕ = 0, thus emf induced will be zero.
  • #22-c
    the net charge displaced in part
    (a).
    Ans : The current flowing in the coil is given by
    `` i=\frac{e}{R}=\frac{2\times {10}^{-3}}{4}=\frac{1}{2}\times {10}^{-3}``
    = 0.5 × 10-3 = 5 × 10-4 A
    Hence, the net charge is given by
    Q = idt = 5 × 10-4 × `` \frac{1}{10}``
    = 5 × 10-5 C
    Page No 307: