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  • Solve this problem Two short bar magnets of length 1 cm 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 magn

Two short bar magnets of length 1 cm 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 20.0 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 x 10-5  Wb/m2 )

 

  • Option 1)

    5.80\times 10^{-4}\; Wb/m^{2}

  • Option 2)

    3.6\times 10^{-5}\; Wb/m^{2}

  • Option 3)

    2.56\times 10^{-4}\; Wb/m^{2}

  • Option 4)

    3.50\times 10^{-4}\; Wb/m^{2}

 

Answers (1)

best_answer

As we discussed in concept 

If a<B_{e}= \frac{\mu _{o}}{4\pi }\frac{M}{r^{3}}

-

 

 M_{1}= 1.00 Am^{2},M_{2}= 1.00 Am^{2}

r= \frac{20}{2}cm= 0.1m

B_{net}= B_{1}+B_{2}+B_{4}

B_{net}= \frac{\mu _{0}}{4\pi}\frac{\left ( M_{1}+M_{2} \right )}{r^{3}}+B_{4}

B_{net}= \frac{10^{-7}\left ( 1.2+1 \right )}{0.1^{3}}+3.6\times 10^{-5}= 2.56\times 10^{-4} \frac{Wb}{m^{2}}


Option 1)

5.80\times 10^{-4}\; Wb/m^{2}

This solution is incorrect 

Option 2)

3.6\times 10^{-5}\; Wb/m^{2}

This solution is incorrect 

Option 3)

2.56\times 10^{-4}\; Wb/m^{2}

This solution is correct 

Option 4)

3.50\times 10^{-4}\; Wb/m^{2}

This solution is incorrect 

Posted by

divya.saini

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