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Karnataka Board PUCPUC Science Class 11

The Magnetic Field at the Origin Due to a Current Element I D → L Placed at a Position → R is - Physics

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Question

The magnetic field at the origin due to a current element  \[i d \vec{l}\] placed at a position \[\vec{r}\] is 

(a)\[\frac{\mu_0 i}{4\pi}\frac{d \vec{l} \times \vec{r}}{r^3}\]

(b) \[- \frac{\mu_0 i}{4\pi}\frac{\vec{r} \times d \vec{l}}{r^3}\]

(c) \[\frac{\mu_0 i}{4\pi}\frac{\vec{r} \times d \vec{l}}{r^3}\]

(d) \[- \frac{\mu_0 i}{4\pi}\frac{d \vec{l} \times \vec{r}}{r^3}\] 

Short/Brief Note
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Solution

(a) \[\frac{\mu_0 i}{4\pi}\frac{d \vec{l} \times \vec{r}}{r^3}\]

(b) \[- \frac{\mu_0 i}{4\pi}\frac{\vec{r} \times d \vec{l}}{r^3}\]

The magnetic field at the origin due to current element \[i d \vec{l}\]   placed at a position  \[\vec{r}\]  is given by \[d \vec{B} = \frac{\mu_o}{4\pi}i\frac{\vec{r} \times d \vec{l}}{r^3}\]
According to the cross product property,

\[\vec{A} \times \vec{B} = - \vec{B} \times \vec{A} \]
\[ \Rightarrow d \vec{B} = - \frac{\mu_o}{4\pi}i\frac{\vec{r} \times \vec{dl}}{r^3}\] 

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Chapter 13: Magnetic Field due to a Current - MCQ [Page 249]

APPEARS IN

HC Verma Concepts of Physics Vol. 2 [English] Class 11 and 12
Chapter 13 Magnetic Field due to a Current
MCQ | Q 1 | Page 249
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