Question

Two insulated long straight wires, each carrying \( 2.0 \ \text{A} \) current are kept along \( xx' \) and \( yy' \) axes as shown in the figure. Find the magnitude and direction of the resultant magnetic field at point \( P(4\,\text{m},\,5\,\text{m}) \).

Answer 1

At point \( P(4\,\text{m},\,5\,\text{m}) \):
- The distance from the wire along the y-axis (perpendicular distance) is \( x = 4\,\text{m} \).
- The distance from the wire along the x-axis (perpendicular distance) is \( y = 5\,\text{m} \).
The magnetic field due to a long straight wire at a perpendicular distance \( r \) is given by:\[B = \frac{\mu_0 I}{2\pi r}\]For the y-axis wire (current \( I = 2 \) A) at point P:\[B_1 = \frac{\mu_0 \cdot 2}{2\pi \cdot 4} = \frac{\mu_0}{4\pi} \ \text{(directed into the page)}\]For the x-axis wire (current \( I = 2 \) A) at point P:\[B_2 = \frac{\mu_0 \cdot 2}{2\pi \cdot 5} = \frac{\mu_0}{5\pi} \ \text{(directed into the page)}\]Since both magnetic fields are directed into the page, the resultant magnetic field \( B \) is their vector sum:\[B = \sqrt{B_1^2 + B_2^2}= \frac{\mu_0}{\pi} \sqrt{\left(\frac{1}{4}\right)^2 + \left(\frac{1}{5}\right)^2}= \frac{\mu_0}{\pi} \sqrt{\frac{1}{16} + \frac{1}{25}}= \frac{\mu_0}{\pi} \sqrt{\frac{41}{400}} = \frac{\mu_0}{\pi} \cdot \frac{\sqrt{41}}{20}\]\[B = \frac{\mu_0 \sqrt{41}}{20\pi}\]Direction:
- \( B_1 \) is due to the vertical wire, directed along the negative z-axis (into the page).
- \( B_2 \) is due to the horizontal wire, also directed along the negative z-axis (into the page).
Therefore, the resultant magnetic field is also directed into the page (–z direction).