A bar magnet of length 14 cm is placed in the magnetic meridian with its north pole pointing towards the geographic north pole. A neutral point is obtained at a distance of 18 cm from the centre of the magnet. If BH = 0.4G, the magnetic moment of the magnet is 1G = 10-4T

Option 1 -

2.880 x103 JT-1 

Option 2 -

2.880J T-1

Option 3 -

2.880 x102 J T-1 

Option 4 -

28.80J T-1

0 4 Views | Posted a month ago
Asked by Shiksha User

  • 1 Answer

  • R

    Answered by

    Raj Pandey | Contributor-Level 9

    a month ago
    Correct Option - 4


    Detailed Solution:

    Kindly consider the following Image

     

Similar Questions for you

V
Vishal Baghel

I010=I0cos? 2? \frac {I_0} {10} = I_0 \cos^2 \theta cos? ? =110=0.31<12which is 0.707\cos \theta = \frac {1} {\sqrt {10} = 0.31 < \frac {1} {\sqrt {2} \quad \text {which is 0.707}

So,

? >45? and90? ? ? <45? \theta > 45^\circ \quad \text {and} \quad 90^\circ - \theta < 45^\circ

so only one option is correct i.e. 18.4°

Angle rotated should be

=90? ? 71.6? =18.4? = 90^\circ - 71.6^\circ = 18.4^\circ

Answer: (A) 18.4°

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Kindly go through the solution

 

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