Question

Four long straight thin wires are held vertically at the corners A, B, C and D of a square of side \( a \), kept on a table and carry equal current \( I \). The wire at A carries current in upward direction whereas the current in the remaining wires flows in downward direction. The net magnetic field at the centre of the square will have the magnitude:

• A. \( \dfrac{\mu_0 I}{\pi a} \) and directed along OC
• B. \( \dfrac{\mu_0 I}{\pi a \sqrt{2}} \) and directed along OD
• C. \( \dfrac{\mu_0 I \sqrt{2}}{\pi a} \) and directed along OB
• D. \( \dfrac{2\mu_0 I}{\pi a} \) and directed along OA

Hint:

For multiple long wires at square corners: - Use symmetry first. - Distance from centre to corner = \( a/\sqrt{2} \). - Always apply right-hand thumb rule for direction.

Answer 1

The Correct Option is C

To find the net magnetic field at the centre of the square due to the four wires carrying current, we can use the Biot-Savart Law, which states that the magnetic field, \( \mathbf{B} \), due to a long straight current-carrying conductor is given by:

\(\mathbf{B} = \frac{\mu_0 I}{2 \pi r}\)

where:

• \( \mu_0 \) is the permeability of free space.
• \( I \) is the current.
• \( r \) is the perpendicular distance from the wire to the point where the magnetic field is being calculated.

 

In this problem, the wires are placed at the corners of a square of side length \( a \). The distance from each corner of the square to its center, which is also the point where we want to calculate the net magnetic field, can be found using the Pythagorean theorem. The diagonal of the square is \(\sqrt{a^2 + a^2} = a\sqrt{2}\). Therefore, the distance from each wire to the center is half the diagonal, \(\frac{a\sqrt{2}}{2}\).

Now, let's consider the contribution of the magnetic field from each wire:

1. The wire at corner A carries current upwards. The direction of the magnetic field it produces at the center, using the right-hand rule, will be circular and directed towards C along diagonal OC.
2. The wires at corners B, C, and D carry current downwards. Using the right-hand rule, each will produce a magnetic field at the center directed radially inward towards each respective wire. Thus:
   • Wire at B produces a magnetic field directed towards A, i.e., along diagonal OB.
   • Wire at D produces a magnetic field directed towards C, i.e., along diagonal OD.

The vectors from wires B, C, and D cancel each other out evenly along diagonally opposite directions. However, wire A has a component along OB, which does not get canceled out. The net magnetic field thus has a direction along OB.

To determine the magnitude of the net magnetic field at the center:

\(B_{\text{total}} = 2 \times \frac{\mu_0 I}{\pi \left(\frac{a}{\sqrt{2}}\right)} \quad = \quad \frac{\mu_0 I \sqrt{2}}{\pi a}\)

Therefore, the net magnetic field at the center of the square has a magnitude of \(\frac{\mu_0 I \sqrt{2}}{\pi a}\) and is directed along OB. This matches the correct option:

Correct Answer: \( \dfrac{\mu_0 I \sqrt{2}}{\pi a} \) and directed along OB