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NEET-XII-Physics

36: Permanent Magnets

with Solutions - page 5
Qstn# iv-25 Prvs-Qstn
  • #25
    A short magnet makes 40 oscillations per minute when used in an oscillation magnetometer at a place where the earth’s horizontal magnetic field is 25 μT. Another short magnet of magnetic moment 1.6 A m2 is placed 20 cm east of the oscillating magnet. Find the new frequency of oscillation if the magnet has its north pole (a) towards north and (b) towards south.
    Ans : Here,
    Frequency of oscillations, `` \nu ``1 = 40 oscillations/min
    Earth's horizontal magnetic field, BH = 25 μT
    Magnetic moment of the second magnet, M = 1.6 A-m2
    Distance at which another short magnet is placed, d = 20 cm = 0.2 m (a) For the north pole of the short magnet facing the north, frequency `` \left({v}_{1}\right)`` is given by
    `` {v}_{1}=\frac{1}{2\,\mathrm{\,\pi \,}}\sqrt{\frac{M{B}_{H}}{I}}``
    `` ``
    `` ``
    Here,
    M = Magnetic moment of the magnet
    I = Moment of inertia
    BH = Horizontal component of the magnetic field
    Now, let B be the magnetic field due to the short magnet.
    When the north pole of the second magnet faces the north pole of the first magnet, the effective magnetic field `` \left({B}_{eff}\right)`` is given by
    Beffective = BH `` -`` B
    The new frequency of oscillations `` \left({v}_{2}\right)`` on placing the second magnet is given by
    `` {v}_{\mathit{2}}\mathit{=}\frac{1}{2\,\mathrm{\,\pi \,}}\sqrt{\frac{M\left({B}_{H}\mathit{-}B\right)}{\mathit{1}}}``
    `` ``
    The magnetic field produced by the short magnet `` \left(B\right)`` is given by
    `` B=\frac{{\,\mathrm{\,\mu \,}}_{0}}{4\,\mathrm{\,\pi \,}}\frac{m}{{d}^{\mathit{3}}}``
    `` \Rightarrow B=\frac{{10}^{-7}\times 1.6}{8\times {10}^{-3}}=20\,\mathrm{\,\mu T\,}``
    `` ``
    Since the frequency is proportional to the magnetic field,
    `` \frac{{v}_{\mathit{1}}}{{v}_{\mathit{2}}}=\sqrt{\frac{{B}_{H}}{{B}_{H}\mathit{-}B}}``
    `` \Rightarrow \frac{40}{{v}_{2}}=\sqrt{\frac{25}{5}}``
    `` \Rightarrow \frac{40}{{v}_{2}}=\sqrt{5}``
    `` \Rightarrow {v}_{2}=\frac{40}{\sqrt{5}}=17.88``
    `` =18\,\mathrm{\,oscillations\,}/\,\mathrm{\,min\,}`` (b) For the north pole facing the south,
    `` {v}_{1}=\frac{1}{2\,\mathrm{\,\pi \,}}\sqrt{\frac{M{B}_{H}}{I}}``
    `` \Rightarrow {v}_{2}=\frac{1}{2\,\mathrm{\,\pi \,}}\sqrt{\frac{M\left(B\mathit{+}{B}_{H}\right)}{\mathit{1}}}``
    `` \Rightarrow \frac{{v}_{1}}{{v}_{2}}=\sqrt{\frac{{B}_{H}}{B\mathit{+}{B}_{H}}}``
    `` \Rightarrow \frac{40}{{v}_{2}}=\sqrt{\frac{25}{45}}``
    `` \Rightarrow {v}_{2}=\frac{40}{\sqrt{25/45}}=54\,\mathrm{\,oscillations\,}/\,\mathrm{\,min\,}``
  • #25-a
    towards north and
    Ans : For the north pole of the short magnet facing the north, frequency `` \left({v}_{1}\right)`` is given by
    `` {v}_{1}=\frac{1}{2\,\mathrm{\,\pi \,}}\sqrt{\frac{M{B}_{H}}{I}}``
    `` ``
    `` ``
    Here,
    M = Magnetic moment of the magnet
    I = Moment of inertia
    BH = Horizontal component of the magnetic field
    Now, let B be the magnetic field due to the short magnet.
    When the north pole of the second magnet faces the north pole of the first magnet, the effective magnetic field `` \left({B}_{eff}\right)`` is given by
    Beffective = BH `` -`` B
    The new frequency of oscillations `` \left({v}_{2}\right)`` on placing the second magnet is given by
    `` {v}_{\mathit{2}}\mathit{=}\frac{1}{2\,\mathrm{\,\pi \,}}\sqrt{\frac{M\left({B}_{H}\mathit{-}B\right)}{\mathit{1}}}``
    `` ``
    The magnetic field produced by the short magnet `` \left(B\right)`` is given by
    `` B=\frac{{\,\mathrm{\,\mu \,}}_{0}}{4\,\mathrm{\,\pi \,}}\frac{m}{{d}^{\mathit{3}}}``
    `` \Rightarrow B=\frac{{10}^{-7}\times 1.6}{8\times {10}^{-3}}=20\,\mathrm{\,\mu T\,}``
    `` ``
    Since the frequency is proportional to the magnetic field,
    `` \frac{{v}_{\mathit{1}}}{{v}_{\mathit{2}}}=\sqrt{\frac{{B}_{H}}{{B}_{H}\mathit{-}B}}``
    `` \Rightarrow \frac{40}{{v}_{2}}=\sqrt{\frac{25}{5}}``
    `` \Rightarrow \frac{40}{{v}_{2}}=\sqrt{5}``
    `` \Rightarrow {v}_{2}=\frac{40}{\sqrt{5}}=17.88``
    `` =18\,\mathrm{\,oscillations\,}/\,\mathrm{\,min\,}``
  • #25-b
    towards south.
    Ans : For the north pole facing the south,
    `` {v}_{1}=\frac{1}{2\,\mathrm{\,\pi \,}}\sqrt{\frac{M{B}_{H}}{I}}``
    `` \Rightarrow {v}_{2}=\frac{1}{2\,\mathrm{\,\pi \,}}\sqrt{\frac{M\left(B\mathit{+}{B}_{H}\right)}{\mathit{1}}}``
    `` \Rightarrow \frac{{v}_{1}}{{v}_{2}}=\sqrt{\frac{{B}_{H}}{B\mathit{+}{B}_{H}}}``
    `` \Rightarrow \frac{40}{{v}_{2}}=\sqrt{\frac{25}{45}}``
    `` \Rightarrow {v}_{2}=\frac{40}{\sqrt{25/45}}=54\,\mathrm{\,oscillations\,}/\,\mathrm{\,min\,}``