Question: What is the magnetic field at the common centre of the wire circuit (continuous line) as shown in th...

Question

What is the magnetic field at the common centre of the wire circuit (continuous line) as shown in the figure below

Answer 1

Solution

The problem asks for the magnetic field at the common center (O) of the given wire circuit. The circuit consists of a regular hexagon and a straight wire passing through its center. The arrows indicate the direction of current flow.

Let's analyze the current distribution based on the provided diagram and the assumption that it's a single continuous wire with current entering at one point and exiting at another. The diagram shows current entering at vertex A (top-left) and exiting at vertex D (bottom-right). The current I splits at A into three paths:

Path A-O-D: A straight wire passing through the center O. Let the current in this path be I_1.
Path A-B-C-D: The upper half of the hexagon. Let the current in this path be I_2.
Path A-F-E-D: The lower half of the hexagon. Let the current in this path be I_3.

These three currents recombine at D, so I = I_1 + I_2 + I_3.

Let a be the side length of the regular hexagon. For a regular hexagon, the distance from the center O to any vertex is also a. Let's assume the wire is uniform, so its resistance is proportional to its length. Let R_s be the resistance of one side of the hexagon (length a). The length of the segment AO is a, and the length of OD is a. So, the resistance of AO is R_s and OD is R_s.

The resistances of the three paths are:

Resistance of path AOD: R_{AOD} = R_{AO} + R_{OD} = R_s + R_s = 2R_s.
Resistance of path ABCD: R_{ABCD} = R_{AB} + R_{BC} + R_{CD} = R_s + R_s + R_s = 3R_s.
Resistance of path AFED: R_{AFED} = R_{AF} + R_{FE} + R_{ED} = R_s + R_s + R_s = 3R_s.

Since R_{ABCD} = R_{AFED}, by symmetry, the current will split equally between these two paths: I_2 = I_3. Let V_{AD} be the potential difference between points A and D. V_{AD} = I_1 \cdot R_{AOD} = I_1 \cdot (2R_s) V_{AD} = I_2 \cdot R_{ABCD} = I_2 \cdot (3R_s) Equating the potential differences: I_1 \cdot 2R_s = I_2 \cdot 3R_s 2I_1 = 3I_2 \implies I_1 = \frac{3}{2}I_2.

Now, let's calculate the magnetic field at the center O due to each part of the circuit.

Magnetic field due to the straight wire AOD (carrying current I_1): The point O is on the straight wire AOD. The magnetic field at any point on the axis of a straight current-carrying wire is zero. Therefore, B_{AOD} = 0.
Magnetic field due to the hexagonal paths (carrying currents I_2 and I_3): The magnetic field at the center O due to a straight current-carrying segment of length L at a perpendicular distance r from O, making angles θ_1 and θ_2 with the lines joining O to the ends of the segment, is given by Biot-Savart Law: B = \frac{\mu_0 I}{4\pi r}(\sin\theta_1 + \sin\theta_2)

For a regular hexagon, the perpendicular distance from the center O to the midpoint of any side is r = a \cos(30^\circ) = a \frac{\sqrt{3}}{2}. The angles θ_1 and θ_2 for each segment (e.g., AB) are 30^\circ (since each side subtends an angle of 60^\circ at the center, and the perpendicular bisects this angle). So, for a segment carrying current I_x, the magnitude of the field at O is: B_{segment}(I_x) = \frac{\mu_0 I_x}{4\pi (a\sqrt{3}/2)}(\sin30^\circ + \sin30^\circ) B_{segment}(I_x) = \frac{\mu_0 I_x}{2\pi a\sqrt{3}}(1/2 + 1/2) = \frac{\mu_0 I_x}{2\pi a\sqrt{3}}.

Now let's consider the directions using the right-hand thumb rule:
- Segments A-B, B-C, C-D (carrying I_2): Current I_2 flows from A to B, B to C, and C to D.
  - B_{AB}: Direction is into the page.
  - B_{BC}: Direction is into the page.
  - B_{CD}: Direction is into the page.
  Total field from these three segments: B_{ABC} = 3 \times \frac{\mu_0 I_2}{2\pi a\sqrt{3}} (into the page).
- Segments A-F, F-E, E-D (carrying I_3 = I_2): Current I_2 flows from A to F, F to E, and E to D.
  - B_{AF}: Direction is out of the page.
  - B_{FE}: Direction is out of the page.
  - B_{ED}: Direction is out of the page.
  Total field from these three segments: B_{AFE} = 3 \times \frac{\mu_0 I_2}{2\pi a\sqrt{3}} (out of the page).
The net magnetic field from the hexagonal paths is the vector sum of B_{ABC} and B_{AFE}. Since they have equal magnitudes and opposite directions: B_{hexagon} = B_{ABC} - B_{AFE} = 0.

Therefore, the total magnetic field at the center O is B_{total} = B_{AOD} + B_{hexagon} = 0 + 0 = 0.

The key to this problem is the symmetry of the current distribution. The current splits such that the contributions from the hexagonal paths cancel each other out, and the central wire passes through the point of interest, contributing zero field. This is analogous to the similar question provided, where the field at the center of a symmetric network is zero due to cancellation.