Due to a small magnet, intensity at a distanc | vedclass

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Question

In $C.G.S.$ $B_{	ext {axial }}=9=\frac{2 M}{x^{3}}$.........$(i)$

$\mathrm{B}_{	ext {equatorial }}=\frac{\mathrm{M}}{\left(\frac{\mathrm{x}}{2}

Vedclass · Accepted Answer

$36$

Vedclass · Answer

In $C.G.S.$ $B_{	ext {axial }}=9=\frac{2 M}{x^{3}}$.........$(i)$

$\mathrm{B}_{	ext {equatorial }}=\frac{\mathrm{M}}{\left(\frac{\mathrm{x}}{2}ight)^{32}}=\frac{\mathrm{M}}{\mathrm{x}^{3}}$.........$(ii)$

From equation $(i)$ and $(ii)$, $B$$_{	ext {equatorial }}=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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