NEET-XII-Physics
36: Permanent Magnets
- #2Two long bare magnets are placed with their axes coinciding in such a way that the north pole of the first magnet is 2.0 cm from the south pole of the second. If both the magnets have a pole strength of 10 Am, find the force exerted by one magnet of the other.Ans : Given:
Pole strength = m1 = m2 = 10 Am
Distance between the north pole of the first magnet and the south pole of the second magnet, r = 2 cm = 0.02 m
We know,
Force `` \left(F\right)`` exerted by two magnetic poles on each other is given by
`` F=\frac{{\,\mathrm{\,\mu \,}}_{0}}{4\,\mathrm{\,\pi \,}}\frac{{m}_{1}{m}_{2}}{{r}^{2}}``
`` =\frac{4\,\mathrm{\,\pi \,}\times {10}^{-7}\times {10}^{2}}{4\,\mathrm{\,\pi \,}\times 4\times {10}^{-4}}``
`` =2.5\times {10}^{-2}\,\mathrm{\,N\,}``
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