Question

Two identical circular current loops carrying equal currents are placed with their axes inclined at 45\(^{\circ}\) to each other as shown in the figure. The resultant magnetic field at P is

• A. \( \frac{\mu_0 I}{16\sqrt{2}R} [(\sqrt{2} + 1)\hat{i} + \hat{j}] \)
• B. \( \frac{\mu_0 I}{16\sqrt{2}R} [\sqrt{2}\hat{i} + \hat{j}] \)
• C. \( \frac{\mu_0 I}{16R} [(\sqrt{2} + 1)\hat{i} + \hat{j}] \)
• D. \( \frac{\mu_0 I}{16R} [\sqrt{2}\hat{i} + \hat{j}] \)

Hint:

When combining axial fields, always resolve them into components. If you find a common factor like \( \frac{1}{\sqrt{2}} \), factor it out early to make the expression look like the options provided in the question.

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
We need to find the vector sum of magnetic fields produced by two identical circular loops at a point P that lies on the axes of both loops at a distance of \( \sqrt{3}R \) from their centers.
Step 2: Key Formula or Approach:
The magnetic field at a point on the axis of a circular current-carrying loop of radius \( R \) at a distance \( x \) from the center is:
\[ B = \frac{\mu_0 I R^2}{2(R^2 + x^2)^{3/2}} \] Step 3: Detailed Explanation:
Given:
Distance \( x = \sqrt{3}R \).
Calculating the magnitude of the magnetic field from one loop:
\[ B_0 = \frac{\mu_0 I R^2}{2(R^2 + (\sqrt{3}R)^2)^{3/2}} = \frac{\mu_0 I R^2}{2(R^2 + 3R^2)^{3/2}} = \frac{\mu_0 I R^2}{2(4R^2)^{3/2}} \] \[ B_0 = \frac{\mu_0 I R^2}{2(8R^3)} = \frac{\mu_0 I}{16R} \] Now, let's represent these fields as vectors based on the coordinate system in the figure:
1. The field from the first loop (whose axis is along the X-axis) is:
\( \vec{B}_1 = B_0 \hat{i} \)
2. The field from the second loop (whose axis is inclined at 45\(^{\circ}\) to the X-axis) is:
\( \vec{B}_2 = B_0 (\cos 45^{\circ} \hat{i} + \sin 45^{\circ} \hat{j}) = B_0 \left( \frac{1}{\sqrt{2}}\hat{i} + \frac{1}{\sqrt{2}}\hat{j} \right) \)
The resultant magnetic field \( \vec{B} \) is the vector sum:
\[ \vec{B} = \vec{B}_1 + \vec{B}_2 = B_0 \hat{i} + B_0 \left( \frac{1}{\sqrt{2}}\hat{i} + \frac{1}{\sqrt{2}}\hat{j} \right) \] \[ \vec{B} = B_0 \left[ \left(1 + \frac{1}{\sqrt{2}}\right)\hat{i} + \frac{1}{\sqrt{2}}\hat{j} \right] = \frac{B_0}{\sqrt{2}} [(\sqrt{2} + 1)\hat{i} + \hat{j}] \] Substituting the value of \( B_0 \):
\[ \vec{B} = \frac{\mu_0 I}{16\sqrt{2}R} [(\sqrt{2} + 1)\hat{i} + \hat{j}] \] Step 4: Final Answer:
The resultant magnetic field at P is \( \frac{\mu_0 I}{16\sqrt{2}R} [(\sqrt{2} + 1)\hat{i} + \hat{j}] \), which corresponds to option (1).