question_answer

Two short bar 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 20.0 cm. The value of the resultant horizontal magnetic induction at the mid - point O of the line joining their centre is close to (Horizontal component of earth′s magnetic induction is\[3.6\times {{10}^{-5}}Wb/{{m}^{2}}\])     AIEEE Solevd Paper-2013

A) \[3.6\times {{10}^{-5}}Wb/{{m}^{2}}\]

B) \[2.56\times {{10}^{-4}}Wb/{{m}^{2}}\]

C) \[3.50\times {{10}^{-4}}Wb/{{m}^{2}}\]

D) \[5.80\times {{10}^{-4}}Wb/{{m}^{2}}\]

Correct Answer: B

Solution :

\[B=\frac{{{\mu }_{0}}}{4\pi }\frac{\mu }{{{\left[ {{\left( \frac{L}{2} \right)}^{2}}+{{y}^{2}} \right]}^{3/2}}}\] \[{{B}_{1}}={{10}^{-7}}\times \frac{1.2}{{{\left( {{(0.1)}^{2}}+{{(0.005)}^{2}} \right)}^{3/2}}}\] \[={{10}^{-7}}\frac{(1.2)}{{{10}^{-3}}}=1.2\times {{10}^{-4}}\] \[Wb/{{m}^{2}}\](South to North) \[{{B}_{2}}=1\times {{10}^{-4}}Wb/{{m}^{2}}\] (south to North) \[{{B}_{H}}=3.6\times {{10}^{-5}}Wb/{{m}^{2}}\] (South to North) \[{{B}_{net}}=(0.36+1+1.2)\times {{10}^{-4}}=2.56\times {{10}^{-4}}\] \[Wb/{{m}^{2}}\]