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The conducting triangular loop in this figure carries a current of 10 A. Find the magnetic field intensity (H) at (0, 0, 5) due to side 1 of the loop.

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BHEL Engineer Trainee Electrical 23 Aug 2023 Official Paper
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  1. \(\rm H=-\frac{1}{\pi\sqrt{19}}a_y\) mA/m
  2. \(\rm H=-\frac{1}{\pi\sqrt{27}}a_y\) mA/m
  3. \(\rm H=-\frac{1}{\pi\sqrt{23}}a_y\) mA/m
  4. \(\rm H=-\frac{1}{\pi\sqrt{29}}a_y\) mA/m

Answer (Detailed Solution Below)

Option 4 : \(\rm H=-\frac{1}{\pi\sqrt{29}}a_y\) mA/m
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Detailed Solution

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Concept:

To find the magnetic field intensity \( \vec{H} \) at a point due to a straight current-carrying segment, we use the Biot–Savart law for a finite wire:

\( \vec{H} = \frac{I}{4\pi} \int \frac{d\vec{l} \times \vec{R}}{R^3} \)

Given:

Current \( I = 10 \, A \), Observation point = \( (0, 0, 5) \), Side 1 runs from \( (0, 0, 0) \) to \( (2, 0, 0) \)

Calculation:

Let \( d\vec{l} = dx \, \vec{a_x} \), and \( \vec{R} = -x\vec{a_x} + 5\vec{a_z} \)

\( d\vec{l} \times \vec{R} = dx \cdot (5 \vec{a_y}) \)

So, \( \vec{H} = \frac{10}{4\pi} \int_{0}^{2} \frac{5 \vec{a_y} \, dx}{(x^2 + 25)^{3/2}} \)

Integrating,

\( \int_{0}^{2} \frac{dx}{(x^2 + 25)^{3/2}} = \left[ \frac{x}{25\sqrt{x^2 + 25}} \right]_0^2 = \frac{2}{25 \sqrt{29}} \)

So,

\( \vec{H} = \frac{50 \vec{a_y}}{4\pi} \cdot \frac{2}{25 \sqrt{29}} = \frac{1}{\pi \sqrt{29}} \vec{a_y} \, \text{A/m} \)

Direction is negative due to clockwise loop direction → \( \vec{H} = -\frac{1}{\pi \sqrt{29}} \vec{a_y} \, \text{mA/m} \)

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