Question

calculate the magnetic induction at (3,0) due to small current element of magnitude 0.2 ampere metre placed along y axis at origin (draw a figure and use Biot savart law)​

Answer 1

$B=6.67\times 10^{-9}$ Tesla.

Explanation:

We know , according to Biot savart law, $\vec{B}=\dfrac{\mu_o }{4\pi}\dfrac{I\vec{ds}\times \^{r}}{r^2}.$      ( Equation 1)

Given,

Current element, $I\vec{ds}= 0.2\ \^{j}.$

Displacement vector, $\vec{r}=3\ \^{i}\ .$

Now, $I\vec{ds}\times \^r=0.2 \ \^j\times 3\ \^i=-0.6\ \^k$  ( because, $\^j\times\^i=-\^k$ )

Also, r = 3 meter.

Putting all these values in eqn 1.

We get, $\vec{B}=\dfrac{4\times \pi\times10^{-7} }{4\pi}\dfrac{0.6}{3^2}(-\^k)=-6.67\times 10^{-9}\ \^k\ Tesla.$

Therefore,

$B=6.67\times 10^{-9}$ Tesla.

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Magnetostatics

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