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  • I have a doubt, kindly clarify. - At some location on earth the horizontal component of earth's magnetic field is . At this location m - Magnetic Effects of Current and Magnetism - JEE Main

At some location on earth the horizontal component of earth's magnetic field is 18\times 10^{-6}T. At this location magnetic of length 0.12 m and pole strength 1.8 Am is suspended from its mid -point using a thread, it makes 450 angle with horizontal in equilibrium . To keep this needle horizontal, the vertical force that should be applied at one of its ends is:

 

  • Option 1)

    1.3\times 10^{-5} N^{}

  • Option 2)

    6.5\times 10^{-5} N

  • Option 3)

    1.6\times 10^{-5} N

  • Option 4)

    1.8\times 10^{-5} N

Answers (1)

best_answer

 

Torque -

\underset{T}{\rightarrow}=\underset{M}{\rightarrow}\times \underset{B}{\rightarrow}=M=NiA

T=MB\sin \theta =NBiA\sin \theta

 

- wherein

M - magnetic moment 

 

 

Resultant Magnetic Field due to earth -

B_{H}=B\cos \phi

 

 

B_{V}=B\sin \phi

B=\sqrt{B_{H}^{2}+B_{V}^{2}}

-

 

MB\sin 45 = F\frac{l}{2}\sin 45

F = 2MB


Option 1)

1.3\times 10^{-5} N^{}

Option 2)

6.5\times 10^{-5} N

Option 3)

1.6\times 10^{-5} N

Option 4)

1.8\times 10^{-5} N

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