NEET-XII-Physics
36: Permanent Magnets
- #6Two short magnets of equal dipole moments M are fastened perpendicularly at their centre (figure 36-Q1). The magnitude of the magnetic field at a distance d from the centre on the bisector of the right angle is
(a)
μ04πMd3
(b)
μ04π2 Md3
(c)
μ04π22Md3
(d)
μ04π2Md3Figure
digAnsr: cAns : (c) `` \frac{{\mu }_{0}}{4\pi }\frac{2\sqrt{2}M}{{d}^{3}}``
Magnetic field (B1) due to the short dipole A of dipole moment M at an axial point is given by,
`` {\stackrel{\to }{B}}_{1}=\frac{{\mu }_{0}}{4\,\mathrm{\,\pi \,}}\frac{2M}{{d}^{3}}...\left(1\right)``
Magnetic field (B2) due to the short dipole B of dipole moment M at an axial point is given by,
`` {\stackrel{\to }{B}}_{2}=\frac{{\mu }_{0}}{4\,\mathrm{\,\pi \,}}\frac{2M}{{d}^{3}}...\left(2\right)``
Resultant magnetic field (B) will be,
B = `` \sqrt{{{B}_{1}}^{2}+{{B}_{2}}^{2}}``
B = `` \frac{{\mu }_{0}}{4\,\mathrm{\,\pi \,}}\frac{2\sqrt{2}M}{{d}^{3}}``
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