The magnetic field at the origin due to a cur | vedclass

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(B) According to the Biot-Savart Law,the magnetic field $d\vec{B}$ at a point $P$ due to a current element $i \, d\vec{l}$ located at a position vecto

Vedclass · Accepted Answer

$(ii), (iii)$

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(B) According to the Biot-Savart Law,the magnetic field $d\vec{B}$ at a point $P$ due to a current element $i \, d\vec{l}$ located at a position vector $\vec{r}'$ relative to the source is given by $d\vec{B} = \frac{\mu_0}{4\pi} \frac{i \, d\vec{l} 	imes \vec{r}_{rel}}{r_{rel}^3}$,where $\vec{r}_{rel}$ is the vector from the source to the point of observation.Here,the current element is at position $\vec{r}$ and we need the field at the origin $(0,0,0)$. Thus,the vector from the source to the origin is $\vec{r}_{rel} = \vec{0} - \vec{r} = -\vec{r}$.Substituting this into the formula:$d\vec{B} = \frac{\mu_0 i}{4\pi} \frac{d\vec{l} 	imes (-\vec{r})}{r^3} = -\frac{\mu_0 i}{4\pi} \frac{d\vec{l} 	imes \vec{r}}{r^3}$. This matches expression $(ii)$.Using the property of cross products,$d\vec{l} 	imes \vec{r} = -(\vec{r} 	imes d\vec{l})$.Substituting this into the expression for $d\vec{B}$:$d\vec{B} = -\frac{\mu_0 i}{4\pi} \frac{-(\vec{r} 	imes d\vec{l})}{r^3} = \frac{\mu_0 i}{4\pi} \frac{\vec{r} 	imes d\vec{l}}{r^3}$. This matches expression $(iii)$.Therefore,expressions $(ii)$ and $(iii)$ are correct.

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