A magnet of length 10 text{ cm} and magnetic | vedclass

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(D) The magnet is placed along $AB$. The point $C$ is at the equatorial position relative to the center $O$ of the magnet.Length of magnet $2l = 10

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$10^{-4} 	ext{ T}$

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(D) The magnet is placed along $AB$. The point $C$ is at the equatorial position relative to the center $O$ of the magnet.Length of magnet $2l = 10 	ext{ cm} = 0.1 	ext{ m}$, so $l = 0.05 	ext{ m}$.Magnetic moment $M = 1 	ext{ Am}^2$.The distance $OC$ is the height of the equilateral triangle with side $a = 10 	ext{ cm} = 0.1 	ext{ m}$.$OC = \sqrt{a^2 - (a/2)^2} = \sqrt{0.1^2 - 0.05^2} = \sqrt{0.01 - 0.0025} = \sqrt{0.0075} = \sqrt{75} 	imes 10^{-2} 	ext{ m} = 5\sqrt{3} 	imes 10^{-2} 	ext{ m} \approx 0.0866 	ext{ m}$.The magnetic field at an equatorial point is given by $B = \frac{\mu_0}{4\pi} \frac{M}{(OC^2 + l^2)^{3/2}}$.Since $OC \gg l$ is not strictly true here, we use the general formula $B = \frac{\mu_0}{4\pi} \frac{M}{(r^2 + l^2)^{3/2}}$ where $r = OC$.$B = 10^{-7} 	imes \frac{1}{((0.0866)^2 + (0.05)^2)^{3/2}} = 10^{-7} 	imes \frac{1}{(0.0075 + 0.0025)^{3/2}} = 10^{-7} 	imes \frac{1}{(0.01)^{3/2}} = 10^{-7} 	imes \frac{1}{(10^{-2})^{3/2}} = 10^{-7} 	imes \frac{1}{10^{-3}} = 10^{-4} 	ext{ T}$.

$A$ magnet of length $10 \text{ cm}$ and magnetic moment $1 \text{ Am}^2$ is placed along side $AB$ of an equilateral triangle $ABC$. If the length of the side $AB$ is $10 \text{ cm}$, find the magnetic induction at point $C$. (Given $\mu_0 = 4\pi \times 10^{-7} \text{ Hm}^{-1}$)

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