NEET-XII-Physics
36: Permanent Magnets
- #7A bar magnet has a length of 8 cm. The magnetic field at a point at a distance 3 cm from the centre in the broadside-on position is found to be 4 × 10-6 T. Find the pole strength of the magnet.Ans : Given:
Length of the magnet, 2l = 8 cm = 8 `` \times `` 10`` -2`` m
Distance of the observation point from the centre of the dipole, d = 3 cm
Magnetic field in the broadside-on position, B = 4 × 10-6 T
The magnetic field due to the dipole on the equatorial point `` \left(B\right)``is given by
`` B=\frac{{\,\mathrm{\,\mu \,}}_{0}m2l}{4\,\mathrm{\,\pi \,}{\left({d}^{2}+{l}^{2}\right)}^{3/2}}``
`` ``
Here, m is the pole strength of the magnet.
On substituting the respective values, we get
`` 4\times {10}^{-6}=\frac{{10}^{-7}m\times 8\times {10}^{-2}}{{\left(9\times {10}^{-4}+16\times {10}^{-4}\right)}^{3/2}}``
`` \Rightarrow 4\times {10}^{-6}=\frac{m\times 8\times {10}^{-9}}{{\left(25\right)}^{3/2}\times {\left({10}^{-4}\right)}^{3/2}}``
`` \Rightarrow m=\frac{4\times {10}^{-6}\times 125\times {10}^{-6}}{8\times {10}^{-9}}``
`` \Rightarrow m=6.25\times {10}^{-2}\,\mathrm{\,A\,}-\,\mathrm{\,m\,}``
Thus, the pole strength of the magnet is 6.25`` \times ``10`` -2`` A-m.
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