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 \( P(4\,\text{m},\,5\,\text{m}) \):
- Distance from wire along y-axis = \( x = 4\,\text{m} \)
- Distance from wire along x-axis = \( y = 5\,\text{m} \)
Magnetic field due to a long straight wire at a perpendicular distance \( r \) is: \[ B = \frac{\mu_0 I}{2\pi r} \] So, Magnetic field due to y-axis wire at \( P \): \[ B_1 = \frac{\mu_0 \cdot 2}{2\pi \cdot 4} = \frac{\mu_0}{4\pi} \ \text{(into the page)} \] Magnetic field due to x-axis wire at \( P \): \[ B_2 = \frac{\mu_0 \cdot 2}{2\pi \cdot 5} = \frac{\mu_0}{5\pi} \ \text{(into the page)} \] Since both are into the page, resultant magnetic field \( B \) is the 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 vertical wire, directed along negative z-axis. - \( B_2 \) is due to horizontal wire, also directed along negative z-axis. So the resultant magnetic field is also into the page (–z direction).