Question 27 of 57

Two short bar magnets of length $1cm$ each have magnetic moments $1.20A{m}^{2}$ and $1.00A{m}^{2}$ respectively. They are placed on a horizontal table parallel to each other with their $\text{N}$ poles pointing towards the south. They have a common magnetic equator and are separated by a distance of $20.0cm$. The value of the resultant horizontal magnetic induction at the mid-point $\text{O}$ 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$3.50\times {10}^{-4}Wb{m}^{-2}$B$5.80\times {10}^{-4}Wb{m}^{-2}$C$3.6\times {10}^{-5}Wb{m}^{-2}$D$2.56\times {10}^{-4}Wb{m}^{-2}$

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Answer: (D) $2.56\times {10}^{-4}Wb{m}^{-2}$

${\text{B}}_{\text{net}}={\text{B}}_{1}+{\text{B}}_{2}+{\text{B}}_{H}$

$=\dfrac{{\mu }_{0}}{4\pi }\dfrac{{\text{M}}_{1}}{{\text{x}}^{3}}+\dfrac{{\mu }_{0}}{4\pi }\dfrac{{\text{M}}_{2}}{{\text{x}}^{3}}+{\text{B}}_{\text{H}}$

$=\dfrac{{{10}^{-7}(1.2+1)}}{{(0.1)}^{3}}+3.6\times {10}^{-5}$

$=\text{2.56}\times {\text{10}}^{-4}\text{Wb}/{\text{m}}^{2}$