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NEET-XII-Physics

38: Electromagnetic Induction

with Solutions - page 9
Qstn# iv-38-a Prvs-QstnNext-Qstn
  • #38-a
    the current in the loop at an instant when the speed of the wire PQ is v, (b) the acceleration of the wire at this instant, (c) the velocity v as a functions of x and (d) the maximum distance the wire will move.
    Figure (b) the acceleration of the wire at this instant, (c) the velocity v as a functions of x and (d) the maximum distance the wire will move.
    Figure
    Ans : When wire PQ is moving with a speed v, the emf induced across it is given by
    e = Blv
    Total resistance of the circuit = r + R
    ∴ Current in the circuit, i = `` \frac{Blv}{r+R}`` (b) Force acting on the wire at the given instant, F = ilB
    On substituting the value of i from above, we get
    `` F=\frac{\left(Blv\right)\left(lB\right)}{(R+r)}=\frac{{B}^{2}{l}^{2}v}{R+r}``
    Acceleration of the wire is given by
    a`` =\frac{{B}^{2}{l}^{2}v}{m(R+r)}`` (c) Velocity can be expressed as:
    v = v0 + at = `` {v}_{0}-\frac{{B}^{2}{l}^{2}v}{m(R+r)}t`` (As force is opposite to velocity)
    Velocity as the function of x is given by
    `` v={v}_{0}-\frac{{B}^{2}{l}^{2}x}{m(R+r)}``
    `` \left(d\right)a=v\frac{dv}{dx}=\frac{{B}^{2}{l}^{2}v}{m(R+r)}``
    `` dx=\frac{m(R+r)}{{B}^{2}{l}^{2}}dv``
    On integrating both sides, we get
    `` x=\frac{m(R+r){v}_{0}}{{B}^{2}{l}^{2}}``
    Page No 309: (b) Force acting on the wire at the given instant, F = ilB
    On substituting the value of i from above, we get
    `` F=\frac{\left(Blv\right)\left(lB\right)}{(R+r)}=\frac{{B}^{2}{l}^{2}v}{R+r}``
    Acceleration of the wire is given by
    a`` =\frac{{B}^{2}{l}^{2}v}{m(R+r)}`` (c) Velocity can be expressed as:
    v = v0 + at = `` {v}_{0}-\frac{{B}^{2}{l}^{2}v}{m(R+r)}t`` (As force is opposite to velocity)
    Velocity as the function of x is given by
    `` v={v}_{0}-\frac{{B}^{2}{l}^{2}x}{m(R+r)}``
    `` \left(d\right)a=v\frac{dv}{dx}=\frac{{B}^{2}{l}^{2}v}{m(R+r)}``
    `` dx=\frac{m(R+r)}{{B}^{2}{l}^{2}}dv``
    On integrating both sides, we get
    `` x=\frac{m(R+r){v}_{0}}{{B}^{2}{l}^{2}}``
    Page No 309:
  • #38-b
    the acceleration of the wire at this instant,
    Ans : Force acting on the wire at the given instant, F = ilB
    On substituting the value of i from above, we get
    `` F=\frac{\left(Blv\right)\left(lB\right)}{(R+r)}=\frac{{B}^{2}{l}^{2}v}{R+r}``
    Acceleration of the wire is given by
    a`` =\frac{{B}^{2}{l}^{2}v}{m(R+r)}``
  • #38-c
    the velocity v as a functions of x and (d) the maximum distance the wire will move.
    Figure
    Ans : Velocity can be expressed as:
    v = v0 + at = `` {v}_{0}-\frac{{B}^{2}{l}^{2}v}{m(R+r)}t`` (As force is opposite to velocity)
    Velocity as the function of x is given by
    `` v={v}_{0}-\frac{{B}^{2}{l}^{2}x}{m(R+r)}``
    `` \left(d\right)a=v\frac{dv}{dx}=\frac{{B}^{2}{l}^{2}v}{m(R+r)}``
    `` dx=\frac{m(R+r)}{{B}^{2}{l}^{2}}dv``
    On integrating both sides, we get
    `` x=\frac{m(R+r){v}_{0}}{{B}^{2}{l}^{2}}``
    Page No 309: