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A magnet of length $10 \mathrm{~cm}$ and magnetic moment $1 \mathrm{Am}^2$ is placed along side $A B$ of an equilateral triangle $A B C$. If the length of the side $A B$ is $10 \mathrm{~cm}$. The magnetic induction at the point $C$ is $\left(\mu_0=4 \pi \times 10^{-7} \mathrm{Hm}^{-1}\right)$
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The correct answer is:
$10^{-4} \mathrm{~T}$
$l=10 \mathrm{~cm}, M=1 \mathrm{Am}^2$
Magnetic induction at point ${ }^{+} \mathrm{C}$
(In equatorial position),
$\begin{aligned} B & =\frac{\mu_a}{4 \pi} \frac{M}{(C O)^2}=10^{-7} \times \frac{1}{\left(\sqrt{10^2-5^2}\right)^3} \\ & =\frac{10^{-7} \times 1}{(\sqrt{75})^3}=\frac{10^{-7}}{\left(5 \sqrt{3} \times 10^{-2}\right)^3} \\ & =\frac{10^{-7}}{125 \times 3 \sqrt{3} \times 10^{-6}}=\frac{10^{-1}}{375 \times 1.732} \\ & =0.000154=1.54 \times 10^{-4} \mathrm{~T} \\ & =10^{-4} \mathrm{~T}\end{aligned}$
Magnetic induction at point ${ }^{+} \mathrm{C}$
(In equatorial position),
$\begin{aligned} B & =\frac{\mu_a}{4 \pi} \frac{M}{(C O)^2}=10^{-7} \times \frac{1}{\left(\sqrt{10^2-5^2}\right)^3} \\ & =\frac{10^{-7} \times 1}{(\sqrt{75})^3}=\frac{10^{-7}}{\left(5 \sqrt{3} \times 10^{-2}\right)^3} \\ & =\frac{10^{-7}}{125 \times 3 \sqrt{3} \times 10^{-6}}=\frac{10^{-1}}{375 \times 1.732} \\ & =0.000154=1.54 \times 10^{-4} \mathrm{~T} \\ & =10^{-4} \mathrm{~T}\end{aligned}$
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