Question

Consider a 10-cm long piece of a wire which carries a current of 10 A. Find the magnitude of the magnetic field due to the piece at a point which makes an equilateral triangle with the ends of the piece.

Answer 1

Consider a 10-cm long piece of a wire which carries a current of 10 A. The magnitude of the magnetic field due to the piece at a point which makes an equilateral triangle with the ends of the piece is $11.56 \times 10^{-6} T$

Explanation:

Step 1:

Let AB be 10 cm length wire, and let P be the required point.

Given data  :

Current Magnitude, i = 10 A

The angles of points A and B, point P, are

$\theta_{1}=30^{\circ} \text { and } \theta_{2}=30^{\circ}$

∴ Divide the point from wire, $\mathrm{d}=\sqrt{\frac{3}{2}} \mathrm{a}=\frac{\sqrt{3}}{2} \times 10=5 / \sqrt{3} \mathrm{cm}$

Step 2:

Therefore, the magnetic field is supplied by current in the wire

$B=\frac{\mu_{0} i}{4 \pi d}\left(\sin \theta_{1}+\sin \theta_{2}\right)$

On substituting the values, we get

   $=\frac{10^{-7} \times 10}{5 \sqrt{3} \times 10^{-2}}\left(\frac{1}{2}+\frac{1}{2}\right)$

     $=\frac{10^{-5} \times 10}{5 \sqrt{3}} \times 1$

    $=\frac{10^{-5} \times 2}{\sqrt{3}}$

    $=\frac{10^{-5} \times 2}{1.73}$

    $\begin{aligned}&=1.156 \times 10^{-5} T\\&=11.56 \times 10^{-6} T\end{aligned}$

Answer 2

$\huge{\boxed{\mathcal\pink{\fcolorbox{red}{yellow}{Answer}}}}$

$11.56 \times {10}^{ - 6} t$

hope it help