Due to a small magnet intensity at a distance | vedclass

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(c) In $C.G.S$. ${B_{axial}} = 9 = \frac{{2M}}{{{x^3}}}$.....$(i)$
${B_{{\rm{equaterial}}}} = \frac{M}{{{{\left( {\frac{x}{2}} \right)}^3}}} = \frac{

Vedclass · Accepted Answer

$36$

Vedclass · Answer

(c) In $C.G.S$. ${B_{axial}} = 9 = \frac{{2M}}{{{x^3}}}$.....$(i)$
${B_{{\rm{equaterial}}}} = \frac{M}{{{{\left( {\frac{x}{2}} \right)}^3}}} = \frac{{8M}}{{{x^3}}}$.....$(ii)$
From equation $(i)$ and $(ii)$ ${B_{{\rm{equaterial}}}} = 36\,$ $Gauss$.

Due to a small magnet intensity at a distance $x$ in the end on position is $9$ $Gauss$. What will be the intensity at a distance $\frac{x}{2}$ on broad side on position..... $Gauss$

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