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NEET-XII-Physics

38: Electromagnetic Induction

with Solutions - page 5
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  • #13
    Figure shows a metallic square frame of edge a in a vertical plane. A uniform magnetic field B exists in the space in a direction perpendicular to the plane of the figure. Two boys pull the opposite corners of the square to deform it into a rhombus. They start pulling the corners at t = 0 and displace the corners at a uniform speed u. (a) Find the induced emf in the frame at the instant when the angles at these corners reduce to 60°. (b) Find the induced current in the frame at this instant if the total resistance of the frame is R. (c) Find the total charge which flows through a side of the frame by the time the square is deformed into a straight line.
    Figure (a) Find the induced emf in the frame at the instant when the angles at these corners reduce to 60°. (b) Find the induced current in the frame at this instant if the total resistance of the frame is R. (c) Find the total charge which flows through a side of the frame by the time the square is deformed into a straight line.
    Figure
    Ans : (a) The effective length of each side is the length that is perpendicular to the velocity of the corners.
    Thus, the effective length of each side is a sin θ.
    Net effective length for four sides = 4 × `` \frac{a}{2}`` = 2a
    ∴ Induced emf = Bvl = 2Bau (b) Current in the frame is given by
    i `` =\frac{e}{R}=\frac{2auB}{R}`` (c) Total charge q, which flows through the sides of the frame, is given by
    `` q=\frac{∆\varphi }{R}``
    Here,
    ΔΦ = Change in the flux
    R = Resistance of the coil
    `` \therefore q=\frac{∆\varphi }{R}``
    `` =\frac{B({a}^{2}-0)}{R}``
    `` =\frac{B{a}^{2}}{R}``
    Page No 307: (a) The effective length of each side is the length that is perpendicular to the velocity of the corners.
    Thus, the effective length of each side is a sin θ.
    Net effective length for four sides = 4 × `` \frac{a}{2}`` = 2a
    ∴ Induced emf = Bvl = 2Bau (b) Current in the frame is given by
    i `` =\frac{e}{R}=\frac{2auB}{R}`` (c) Total charge q, which flows through the sides of the frame, is given by
    `` q=\frac{∆\varphi }{R}``
    Here,
    ΔΦ = Change in the flux
    R = Resistance of the coil
    `` \therefore q=\frac{∆\varphi }{R}``
    `` =\frac{B({a}^{2}-0)}{R}``
    `` =\frac{B{a}^{2}}{R}``
    Page No 307:
  • #13-a
    Find the induced emf in the frame at the instant when the angles at these corners reduce to 60°.
    Ans : The effective length of each side is the length that is perpendicular to the velocity of the corners.
    Thus, the effective length of each side is a sin θ.
    Net effective length for four sides = 4 × `` \frac{a}{2}`` = 2a
    ∴ Induced emf = Bvl = 2Bau
  • #13-b
    Find the induced current in the frame at this instant if the total resistance of the frame is R.
    Ans : Current in the frame is given by
    i `` =\frac{e}{R}=\frac{2auB}{R}``
  • #13-c
    Find the total charge which flows through a side of the frame by the time the square is deformed into a straight line.
    Figure
    Ans : Total charge q, which flows through the sides of the frame, is given by
    `` q=\frac{∆\varphi }{R}``
    Here,
    ΔΦ = Change in the flux
    R = Resistance of the coil
    `` \therefore q=\frac{∆\varphi }{R}``
    `` =\frac{B({a}^{2}-0)}{R}``
    `` =\frac{B{a}^{2}}{R}``
    Page No 307: