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NEET-XII-Physics

36: Permanent Magnets

with Solutions - page 3
Qstn# iv-5-a Prvs-QstnNext-Qstn
  • #5-a
    in end-on position of the dipole and (b) in broadside-on position of the dipole. (b) in broadside-on position of the dipole.
    Ans : If the point is at the end-on position:
    The magnetic field `` \left(B\right)`` on the axial point of the dipole is given by
    `` B=\frac{{\,\mathrm{\,\mu \,}}_{0}}{4\,\mathrm{\,\pi \,}}\frac{2M}{{d}^{3}}``
    `` \,\mathrm{\,Here\,},M\,\mathrm{\,is\,}\,\mathrm{\,the\,}\,\mathrm{\,magnetic\,}\,\mathrm{\,moment\,}\,\mathrm{\,of\,}\,\mathrm{\,the\,}\,\mathrm{\,dipole\,}\,\mathrm{\,that\,}\,\mathrm{\,we\,}\,\mathrm{\,need\,}\,\mathrm{\,to\,}\,\mathrm{\,find\,}\,\mathrm{\,out\,}.``
    `` \therefore 2\times {10}^{-4}=\frac{{10}^{-7}\times 2M}{{\left({10}^{-1}\right)}^{3}}``
    `` \Rightarrow M=\frac{2\times {10}^{-4}\times {10}^{-3}}{{10}^{-7}\times 2}``
    `` \Rightarrow M=1\,\mathrm{\,A\,}-{\,\mathrm{\,m\,}}^{2}`` (b) If the point is at broadside-on position (equatorial position):
    The magnetic field `` \left(B\right)`` is given by
    `` B=\frac{{\,\mathrm{\,\mu \,}}_{0}}{4\,\mathrm{\,\pi \,}}\frac{\mathit{M}}{{\mathit{d}}^{\mathit{3}}}``
    `` \Rightarrow 2\times {10}^{-4}=\frac{{10}^{-7}\times M}{{\left({10}^{-1}\right)}^{3}}``
    `` \Rightarrow M=2\,\mathrm{\,A\,}-{\,\mathrm{\,m\,}}^{2}``
    Page No 277: (b) If the point is at broadside-on position (equatorial position):
    The magnetic field `` \left(B\right)`` is given by
    `` B=\frac{{\,\mathrm{\,\mu \,}}_{0}}{4\,\mathrm{\,\pi \,}}\frac{\mathit{M}}{{\mathit{d}}^{\mathit{3}}}``
    `` \Rightarrow 2\times {10}^{-4}=\frac{{10}^{-7}\times M}{{\left({10}^{-1}\right)}^{3}}``
    `` \Rightarrow M=2\,\mathrm{\,A\,}-{\,\mathrm{\,m\,}}^{2}``
    Page No 277:
  • #5-b
    in broadside-on position of the dipole.
    Ans : If the point is at broadside-on position (equatorial position):
    The magnetic field `` \left(B\right)`` is given by
    `` B=\frac{{\,\mathrm{\,\mu \,}}_{0}}{4\,\mathrm{\,\pi \,}}\frac{\mathit{M}}{{\mathit{d}}^{\mathit{3}}}``
    `` \Rightarrow 2\times {10}^{-4}=\frac{{10}^{-7}\times M}{{\left({10}^{-1}\right)}^{3}}``
    `` \Rightarrow M=2\,\mathrm{\,A\,}-{\,\mathrm{\,m\,}}^{2}``
    Page No 277: