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Two short bar magnets of length $$1\ cm$$ each have magnetic moments $$1.20\ Am^{2}$$ and $$1.00\ Am^{2}$$ respectively. They are placed on a horizontal table parallel to each other with their $$N$$ poles pointing towards the South. They have a common magnetic equator and are separated by a distance of $$20.0\ cm$$. The value of the resultant horizontal magnetic induction at the mid-point $$O$$ of the line joining their centres is close to (Horizontal component of earth's magnetic induction is $$3.6\times 10^{-5}\ Wb/m^{2}$$)


A
2.56×104 Wb/m2
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B
3.50×104 Wb/m2
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C
5.80×104 Wb/m2
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D
3.6×105 Wb/m2
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Solution

The correct option is A $$2.56\times 10^{-4}\ Wb/m^{2}$$
$${B_{net}} = {B_1} + {B_2} + {B_H}$$
$${B_{net}} = \dfrac{{{\mu _0}}}{{4\pi }}\dfrac{{\left( {{M_1} + {M_2}} \right)}}{{{r^3}}} + {B_H}$$
$$ = \dfrac{{{{10}^{ - 7}}\left( {1.2 + 1} \right)}}{{{{\left( {0.1} \right)}^3}}} + 3.6 \times {10^{ - 5}} = 2.56 \times {10^{ - 4}}Wb/{m^3}$$
Hence,
option $$(A)$$ is correct answer.

Physics
NCERT
Standard XII

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