Question
Download Solution PDFThe 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.
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFConcept:
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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