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Two short magnets of length 1 cm each have magnetic moments 1.20 $A m^{2}$
and 1.00 $A m^{2}$ 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 cm. Magnetic induction at the
mid-point ‘o’ of the line joining their centers is close to
$(B_{H} = 3.6 \times 10^{- 5} w b / m^{2})$

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$3.6 \times 10^{- 5} w b / m^{2}$

$2.26 \times 10^{- 4} w b / m^{2}$

$3.50 \times 10^{- 4} w b / m^{2}$

$5.80 \times 10^{- 4} w b / m^{2}$

answer is B.

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Detailed Solution

$B = \frac{μ_{0}}{4 π} \times \frac{M}{d^{3}}$ along equator

$B_{1} = \frac{μ_{0}}{4 π} \times \frac{1.20}{10^{- 3}} = 1.20 \times 10^{- 4} w b / m^{2}$

$B_{2} = \frac{μ_{0}}{4 π} \times \frac{1}{10^{- 3}} = 10^{- 4} w b / m^{2}$

Resultant magnetic field, $B = \sqrt{B_{1}^{2} + B_{2}^{2} + 2 B_{1} B_{2}}$ $(θ = 0^{0})$

$B = \sqrt{1.44 + 1 + 2.4} \times 10^{- 4}$ $= \sqrt{4.84} \times 10^{- 4} = 2.2 \times 10^{- 4} w b / m^{2}$

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