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  • The figure shows a square current carrying loop ABCD of side 10 cm and current  <img alt="\mathrm{i=10 \mathrm{~A}}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmath

The figure shows a square current carrying loop ABCD of side 10 cm and current  \mathrm{i=10 \mathrm{~A}} . The magnetic moment \mathrm{\underset{M}{\rightarrow}} of the loop is

Option: 1

\mathrm{(0.05) (\hat{i}-\sqrt{3 \hat{k}}) A-m^2}


Option: 2

\mathrm{(0.05)(\hat{j}+\hat{k}) A-m^2}


Option: 3

\mathrm{(0.05)(\sqrt{3 \hat{i}}+\hat{k}) A-m^2}


Option: 4

\mathrm{(\hat{i}+\hat{k}) A-m^2}


Answers (1)

best_answer

The magnitude of magnetic moment is
 
\begin{aligned} \mathrm{M}=\mathrm{I} l^2 & =10 \times\left(10 \times 10^{-2}\right)^2 \mathrm{Am}^2 \\ & =10 \times 10^{-2}=0.1 \mathrm{Am}^2 \end{aligned}
 
The normal on the loop is in x-z plane.

It makes 60^{\circ} angle with x-axis.

\begin{aligned} & \therefore \vec{M}=M \cos 60^{\circ} \hat{i}-M \sin 60^{\circ} \hat{j} \\ & \therefore \vec{M}=\frac{M}{2} \hat{i}-\frac{\sqrt{3}}{2} M \hat{j} \\ & \therefore \vec{M}=\frac{0.1}{2}(\hat{i}-\sqrt{3} \hat{j}) \\ & \vec{M}=(0.05)(\hat{i}-\sqrt{3} \hat{j}) A m^2 \end{aligned}

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himanshu.meshram

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