Two short bar magnets of length $$1cm$$ each have magnetic moments $$1.20$$ $$Am2$$ and $$1.00$$ $$Am2$$, 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 $$20.0cm$$. The value of the resultant horizontal magnetic induction at the $$mid-point$$ O of the line joining their centers is close to (horizontal$$ component of earth’s magnetic induction is $$3.6$$×$$10$$−$$5Wb/m2)

Question

Two short bar magnets of length $$1cm$$ each have magnetic moments $$1.20$$ $$Am2$$ and $$1.00$$ $$Am2$$, 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 $$20.0cm$$. The value of the resultant horizontal magnetic induction at the $$mid-point$$ O of the line joining their centers is close to (horizontal$$ component of earth’s magnetic induction is $$3.6$$×$$10$$−$$5Wb/m2)

Infinity Learn · Accepted Answer

The situation is as shown in the figure. As the point O lies on broad side position with respect to both the magnets. Therefore, The net magnetic field at point O is $$Bnet=B1+B2+BH$$ $$Bnet=$$μ$$04$$$$π$$$$M1r3+$$μ$$04$$$$π$$$$M2r3+BH=$$μ$$04$$$$π$$$$r3(M1+M2)+BH$$ Substituting the given values, we get Bnet $$=4$$π×$$10$$−$$74$$π×$$10$$×$$10$$−$$23$$[$$1.2+1$$]$$+3.6$$×$$10$$−$$5=10$$−$$710$$−$$3$$×$$2.2+3.6$$×$$10$$−$$5=2.2$$×$$10$$−$$4+0.36$$×$$10$$−$$4=2.2$$×$$10$$−$$4+0.36$$×$$10$$−$$4=2.56$$×$$10$$−$$4Wb/m2$$

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