A short bar magnet is placed in the magnetic meridian of the earth with north pole pointing north. Neutral points are found at a distance of 30 cm from the magnet on the East - West line, drawn through the middle point of the magnet. The magnetic moment of the magnet in Am2 is close to :

Given \frac{\mu _{0}}{4\pi }=10^{-7}in SI units and

BH=  Horizontal component of earth’s magnetic field   = 3.6 \times 10-5 Tesla.)

 

 

  • Option 1)

    9.7

  • Option 2)

    4.9

  • Option 3)

    19.4

  • Option 4)

    14.6

 

Answers (1)

As we learnt in 

On Equatorial Position -

B_{e}= \frac{\mu _{o}}{4\pi }\frac{M}{\left ( r^{2}+a^{2} \right )^{\frac{3}{2}}}

- wherein

Equatorial position

 

 At 30cm from the magnet on its equotorial plane \overrightarrow{B}_{magnet}= \overrightarrow{B_{H}}

\overrightarrow{B}_{magnet} = \overrightarrow{B_H}\, \, \, \, \because (Neutral \, \, Point)

\frac{\mu _{0}m}{4\pi r^{2}}= 3.6\times 10^{-5}T= \frac{10^{-7}\times m}{\left ( 0.3 \right )^{3}}= 3.6\times 10^{-5}

m=3.6\times 0.07\times 10^{2}= 9.7 Am^{2}


Option 1)

9.7

Correct Option

Option 2)

4.9

Incorrect Option

Option 3)

19.4

Incorrect Option

Option 4)

14.6

Incorrect Option

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