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A dip circle is at right angles to t...

A dip circle is at right angles to the magnetic meridian. What will be the apparent dip ?

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As shown in fig., on loading the upper end of the needle for its rotational, equilibrium,
`vecrxxvacF+vacMxxvacB=0`
`i.e.,MgxxLcos30^(@)=mxx2LxxBsin(45^@-30^@)`
i.e.,`M=2mBsin15^(@)/gcos30^(@)`
Substituting the current data,
`M=2xx200xx0.42xx9.2588/980xx(sqrt(3)//2) = 0.051 g`
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VMC MODULES ENGLISH-MOVING CHARGES & MAGNETISM -Illustration
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