An element dl = dx hat{i} (where,dx = 1 , cm) | vedclass

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(B) Given: Current element $dl = dx \hat{i} = 10^{-2} \, m \hat{i}$,current $i = 10 \, A$,and position vector $\vec{r} = 0.5 \, m \hat{j}$.According t

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$4 	imes 10^{-8} \hat{k} \, T$

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(B) Given: Current element $dl = dx \hat{i} = 10^{-2} \, m \hat{i}$,current $i = 10 \, A$,and position vector $\vec{r} = 0.5 \, m \hat{j}$.According to the Biot-Savart Law,the magnetic field $dB$ is given by:$dB = \frac{\mu_0}{4\pi} \frac{i (d\vec{l} 	imes \vec{r})}{r^3}$Substituting the values:$dB = 10^{-7} 	imes \frac{10 	imes (10^{-2} \hat{i} 	imes 0.5 \hat{j})}{(0.5)^3}$$dB = 10^{-7} 	imes \frac{10 	imes 0.5 	imes 10^{-2} \hat{k}}{0.125}$$dB = 10^{-7} 	imes \frac{0.05 	imes 10^{-2}}{0.125} \hat{k}$$dB = 10^{-7} 	imes \frac{5 	imes 10^{-4}}{12.5 	imes 10^{-2}} \hat{k} = 10^{-7} 	imes 0.4 	imes 10^{-2} \hat{k}$$dB = 4 	imes 10^{-9} 	imes 10 = 4 	imes 10^{-8} \hat{k} \, T$.

An element $dl = dx \hat{i}$ (where,$dx = 1 \, cm$) is placed at the origin and carries a large current $i = 10 \, A$. What is the magnetic field on the $Y$-axis at a distance of $0.5 \, m$?

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