An element overrightarrow{Delta ell} = Delta | vedclass

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

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

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(C) According to the Biot-Savart Law,the magnetic field $d\vec{B}$ due to a current element $I d\vec{\ell}$ is given by: $d\vec{B} = \frac{\mu_0}{4\pi} \frac{I (d\vec{\ell} 	imes \vec{r})}{r^3}$Here,$I = 10 \ A$,$d\vec{\ell} = \Delta x \hat{i} = 10^{-2} \hat{i} \ m$,and the position vector $\vec{r} = 0.5 \hat{j} \ m$.The cross product is $(d\vec{\ell} 	imes \vec{r}) = (10^{-2} \hat{i}) 	imes (0.5 \hat{j}) = 0.5 	imes 10^{-2} (\hat{i} 	imes \hat{j}) = 0.005 \hat{k} \ m^2$.The magnitude of the magnetic field is:$|d\vec{B}| = \frac{\mu_0}{4\pi} \frac{I |d\vec{\ell} 	imes \vec{r}|}{r^3} = 10^{-7} 	imes \frac{10 	imes 0.005}{(0.5)^3}$$|d\vec{B}| = 10^{-7} 	imes \frac{0.05}{0.125} = 10^{-7} 	imes 0.4 = 4 	imes 10^{-8} \ T$.

An element $\overrightarrow{\Delta \ell} = \Delta x \hat{i}$ is placed at the origin and carries a current of $10 \ A$. Find the magnitude of the magnetic field on the $Y$-axis at a distance of $0.5 \ m$,given $\Delta x = 1 \ cm$. (Use $\frac{\mu_0}{4 \pi} = 10^{-7} \ T \cdot m/A$)

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