Question

Two straight long conductors \(\mathrm{AOB}\) and \(\mathrm{COD}\) are perpendicular to each other and carry currents \(\mathrm{I}_{1}\) and \(\mathrm{I}_{2}\). The magnitude of the mag. field at a point " \(\mathrm{P}^{n}\) at a distance " \(\mathrm{a}^{\prime \prime}\) from the point "O" in a direction perpendicular to the plane \(\mathrm{ABCD}\) is (a) \(\left[\left(\mu_{0}\right) /(2 \pi a)\right]\left(I_{1}+I_{2}\right)\) (b) \(\left[\left(\mu_{0}\right) /(2 \pi a)\right]\left(I_{1}-I_{2}\right)\) (c) $\left[\left(\mu_{0}\right) /(2 \pi a)\right]\left(\mathrm{I}_{1}^{2}+\mathrm{I}_{2}^{2}\right)^{1 / 2}$ (d) $\left[\left(\mu_{0}\right) /(2 \pi a)\right]\left[\left(I_{1} I_{2}\right) /\left(I_{1}-I_{2}\right)\right]$

Short answer

The short answer to the question is: \(B = \frac{\mu_{0}}{2\pi a}\sqrt{I_{1}^{2} + I_{2}^{2}}\) This matches option (c) in the given question.

Step-by-step solution

Apply the Biot-Savart law for conductor AOB

Calculate the magnetic field produced by conductor AOB at point P using the Biot-Savart law, which states that the magnetic field dB at a point due to a small current-carrying segment is given by: \(dB = \frac{\mu_{0}}{4\pi}\frac{I\,d\vec{l} \times \vec{r}}{r^{3}}\) where µ₀ = 4π x 10^(-7) Tm/A is the permeability of free space, I is the current through the conductor, d**l** is the small length of the conductor, **r** is the position vector from the current element to the point where we want to calculate the magnetic field, and r is the magnitude of **r**. For conductor AOB, let 'a' be the distance between point O and point P, and let I₁ be the current flowing through the conductor. The magnetic field at point P in the y-direction is given by: \(B_{y1} = \frac{\mu_{0} I_{1}}{2\pi a}\)

Apply the Biot-Savart law for conductor COD

Calculate the magnetic field produced by conductor COD at point P using the Biot-Savart law. Let I₂ be the current flowing through the conductor. The magnetic field at point P in the x-direction is given by: \(B_{x2} = \frac{\mu_{0} I_{2}}{2\pi a}\)

Calculate the net magnetic field vector at point P

Calculate the magnetic field at point P due to both conductors AOB and COD by finding the vector sum of their respective magnetic fields: \(\vec{B} = B_{y1}\hat{y} + B_{x2}\hat{x}\) Substitute the expressions for \(B_{y1}\) and \(B_{x2}\) from Steps 1 and 2: \(\vec{B} = \left(\frac{\mu_{0} I_{1}}{2\pi a}\right)\hat{y} + \left(\frac{\mu_{0} I_{2}}{2\pi a}\right)\hat{x}\)

Calculate the magnitude of the magnetic field at point P

Calculate the magnitude of the magnetic field at point P using the Pythagorean theorem, where the magnitudes of the x and y components are given from Step 3: \(B = \sqrt{(\vec{B}\cdot\vec{B})} = \sqrt{(\frac{\mu_{0} I_{1}}{2\pi a})^{2} + (\frac{\mu_{0} I_{2}}{2\pi a})^{2}}\) Factor out the common term \(\left(\frac{\mu_{0}}{2\pi a}\right)^{2}\): \(B = \frac{\mu_{0}}{2\pi a}\sqrt{I_{1}^{2} + I_{2}^{2}}\) Comparing the expression we found for the magnetic field magnitude with the given options, we see that it matches option (c): \(B = \left[\left(\mu_{0}\right) /(2 \pi a)\right]\left(\mathrm{I}_{1}^{2}+\mathrm{I}_{2}^{2}\right)^{1 / 2}\) Thus, the correct answer is option (c).