A magnetic needle of pole strength $$203$$ Am is pivoted at its centre. Its North pole is pulled eastward by a string. The horizontal force required to produce a deflection of $$300from$$ magnetic meridian (take$$ $$BH=10$$−$$4T)

Question

A magnetic needle of pole strength $$203$$  Am is pivoted at its centre. Its North pole is pulled  eastward by a string. The horizontal force required to produce a deflection of  $$300from$$ magnetic meridian  (take$$  $$BH=10$$−$$4T)

Infinity Learn · Accepted Answer

from the figure   $$cos$$θ$$=yl$$,$$y=lcos$$θ                   Torque due to F is balanced by torque due to BH.⇒Ttension $$=TBH$$⇒$$Fy=MBHsin$$θ⇒Flcosθ$$=m(2l)BHsin$$θ⇒$$F=2mBHtan$$θ   Given  $$m=203Am$$                $$BH=10-4T$$θ$$=30$$°Using, $$F=2mBHtan$$θ⇒$$F=2$$×$$203$$×$$10-4$$×$$13$$ $$=$$ $$4$$×$$10-3$$ N

A magnetic needle of pole strength $20 \sqrt{3}$ Am is pivoted at its centre. Its North pole is pulled eastward by a string. The horizontal force required to produce a deflection of $30^{0}$ from magnetic meridian (take $B_{H} = 10^{- 4} T$ )

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A magnetic needle of pole strength 203 Am is pivoted at its centre. Its North pole is pulled eastward by a string. The horizontal force required to produce a deflection of 300from magnetic meridian (take BH=10−4T)

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A magnetic needle of pole strength $20 \sqrt{3}$ Am is pivoted at its centre. Its North pole is pulled eastward by a string. The horizontal force required to produce a deflection of $30^{0}$ from magnetic meridian (take $B_{H} = 10^{- 4} T$ )