An element Delta l = Delta x hat{i} is placed | vedclass

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(A) According to the Biot-Savart Law, the magnetic field $d\vec{B}$ due to a current element $I d\vec{l}$ is given by:$d\vec{B} = \frac{\mu_0}{4\pi} \

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

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(A) According to the Biot-Savart Law, the magnetic field $d\vec{B}$ due to a current element $I d\vec{l}$ is given by:$d\vec{B} = \frac{\mu_0}{4\pi} \frac{I (d\vec{l} 	imes \vec{r})}{r^3}$Given:$I = 10 \,A$$d\vec{l} = \Delta x \hat{i} = 1 \,cm \cdot \hat{i} = 0.01 \,m \cdot \hat{i}$Position vector $\vec{r} = 0.5 \,m \cdot \hat{j}$Distance $r = 0.5 \,m$Substituting the values:$d\vec{B} = 10^{-7} 	imes \frac{10 	imes (0.01 \hat{i} 	imes 0.5 \hat{j})}{(0.5)^3}$$d\vec{B} = 10^{-7} 	imes \frac{10 	imes 0.005 \hat{k}}{0.125}$$d\vec{B} = 10^{-7} 	imes \frac{0.05}{0.125} \hat{k} = 10^{-7} 	imes 0.4 \hat{k} = 4 	imes 10^{-8} \,T \hat{k}$Thus, the magnitude of the magnetic field is $4 	imes 10^{-8} \,T$.

An element $\Delta l = \Delta x \hat{i}$ is placed at the origin and carries a current $I = 10 \,A$. The magnetic field on the $y$-axis at a distance of $0.5 \,m$ from the element of length $\Delta x = 1 \,cm$ is:

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