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NEET-XII-Physics

38: Electromagnetic Induction

with Solutions - page 9
Qstn# iv-39 Prvs-QstnNext-Qstn
  • #39
    A rectangular frame of wire abcd has dimensions 32 cm × 8.0 cm and a total resistance of 2.0 Ω. It is pulled out of a magnetic field B = 0.020 T by applying a force of 3.2 × 10-5 N (figure). It is found that the frame moves with constant speed. Find (a) this constant speed, (b) the emf induced in the loop, (c) the potential difference between the points a and b and (d) the potential difference between the points c and d.
    Figure (a) this constant speed, (b) the emf induced in the loop, (c) the potential difference between the points a and b and (d) the potential difference between the points c and d.
    Figure
    Ans : Given:
    Total resistance of the frame, R = 2.0 Ω
    Magnetic field, B = 0.020 T
    Dimensions of the frame:
    Length, l = 32 cm = 0.32 m
    Breadth, b = 8 cm = 0.08 m (a) Let the velocity of the frame be v.
    The emf induced in the rectangular frame is given by
    e = Blv
    Current in the coil, `` i=\frac{Blv}{R}``
    The magnetic force on the rectangular frame is given by
    F = ilB = 3.2 × 10-5 N
    On putting the value of i, we get
    `` \frac{{B}^{2}{l}^{2}v}{R}=3.2\times {10}^{-5}``
    `` \Rightarrow \frac{(0.020{)}^{2}\times (0.08{)}^{2}\times v}{2}=3.2\times {10}^{-5}``
    `` \Rightarrow v=\frac{3.2\times {10}^{-5}}{6.4\times {10}^{-3}\times 4\times {10}^{-4}}``
    `` =25\,\mathrm{\,m\,}/\,\mathrm{\,s\,}`` (b) Emf induced in the loop, e = vBl
    ⇒ e = 25 × 0.02 × 0.08
    = 4 × 10-2 V (c) Resistance per unit length is given by
    r`` =\frac{2}{0.8}``
    Ratio of the resistance of part, `` \frac{ad}{cb}=\frac{2\times 0.72}{0.8}=1.8\,\mathrm{\,\Omega \,}``
    `` ``
    `` {V}_{\,\mathrm{\,ab\,}}=iR=\frac{Blv}{2}\times 1.8``
    `` =\frac{0.2\times 0.08\times 25\times 1.8}{2}``
    `` =0.036\,\mathrm{\,V\,}=3.6\times {10}^{-2}\,\mathrm{\,V\,}`` (d) Resistance of cd:
    `` {R}_{\,\mathrm{\,cd\,}}=\frac{2\times 0.8}{0.8}=0.2\,\mathrm{\,\Omega \,}``
    `` V=i{R}_{\,\mathrm{\,cd\,}}=\frac{2\times 0.08\times 25\times 0.2}{2}``
    `` =4\times {10}^{-3}\,\mathrm{\,V\,}``
    Page No 309: (a) Let the velocity of the frame be v.
    The emf induced in the rectangular frame is given by
    e = Blv
    Current in the coil, `` i=\frac{Blv}{R}``
    The magnetic force on the rectangular frame is given by
    F = ilB = 3.2 × 10-5 N
    On putting the value of i, we get
    `` \frac{{B}^{2}{l}^{2}v}{R}=3.2\times {10}^{-5}``
    `` \Rightarrow \frac{(0.020{)}^{2}\times (0.08{)}^{2}\times v}{2}=3.2\times {10}^{-5}``
    `` \Rightarrow v=\frac{3.2\times {10}^{-5}}{6.4\times {10}^{-3}\times 4\times {10}^{-4}}``
    `` =25\,\mathrm{\,m\,}/\,\mathrm{\,s\,}`` (b) Emf induced in the loop, e = vBl
    ⇒ e = 25 × 0.02 × 0.08
    = 4 × 10-2 V (c) Resistance per unit length is given by
    r`` =\frac{2}{0.8}``
    Ratio of the resistance of part, `` \frac{ad}{cb}=\frac{2\times 0.72}{0.8}=1.8\,\mathrm{\,\Omega \,}``
    `` ``
    `` {V}_{\,\mathrm{\,ab\,}}=iR=\frac{Blv}{2}\times 1.8``
    `` =\frac{0.2\times 0.08\times 25\times 1.8}{2}``
    `` =0.036\,\mathrm{\,V\,}=3.6\times {10}^{-2}\,\mathrm{\,V\,}`` (d) Resistance of cd:
    `` {R}_{\,\mathrm{\,cd\,}}=\frac{2\times 0.8}{0.8}=0.2\,\mathrm{\,\Omega \,}``
    `` V=i{R}_{\,\mathrm{\,cd\,}}=\frac{2\times 0.08\times 25\times 0.2}{2}``
    `` =4\times {10}^{-3}\,\mathrm{\,V\,}``
    Page No 309:
  • #39-a
    this constant speed,
    Ans : Let the velocity of the frame be v.
    The emf induced in the rectangular frame is given by
    e = Blv
    Current in the coil, `` i=\frac{Blv}{R}``
    The magnetic force on the rectangular frame is given by
    F = ilB = 3.2 × 10-5 N
    On putting the value of i, we get
    `` \frac{{B}^{2}{l}^{2}v}{R}=3.2\times {10}^{-5}``
    `` \Rightarrow \frac{(0.020{)}^{2}\times (0.08{)}^{2}\times v}{2}=3.2\times {10}^{-5}``
    `` \Rightarrow v=\frac{3.2\times {10}^{-5}}{6.4\times {10}^{-3}\times 4\times {10}^{-4}}``
    `` =25\,\mathrm{\,m\,}/\,\mathrm{\,s\,}``
  • #39-b
    the emf induced in the loop,
    Ans : Emf induced in the loop, e = vBl
    ⇒ e = 25 × 0.02 × 0.08
    = 4 × 10-2 V
  • #39-c
    the potential difference between the points a and b and
    Ans : Resistance per unit length is given by
    r`` =\frac{2}{0.8}``
    Ratio of the resistance of part, `` \frac{ad}{cb}=\frac{2\times 0.72}{0.8}=1.8\,\mathrm{\,\Omega \,}``
    `` ``
    `` {V}_{\,\mathrm{\,ab\,}}=iR=\frac{Blv}{2}\times 1.8``
    `` =\frac{0.2\times 0.08\times 25\times 1.8}{2}``
    `` =0.036\,\mathrm{\,V\,}=3.6\times {10}^{-2}\,\mathrm{\,V\,}``
  • #39-d
    the potential difference between the points c and d.
    Figure
    Ans : Resistance of cd:
    `` {R}_{\,\mathrm{\,cd\,}}=\frac{2\times 0.8}{0.8}=0.2\,\mathrm{\,\Omega \,}``
    `` V=i{R}_{\,\mathrm{\,cd\,}}=\frac{2\times 0.08\times 25\times 0.2}{2}``
    `` =4\times {10}^{-3}\,\mathrm{\,V\,}``
    Page No 309: