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

Hint:

In square-current configurations, first calculate the field due to one wire using \[ B=\frac{\mu_0 I}{2\pi r}, \] then use symmetry and the right-hand thumb rule to determine the direction of the resultant magnetic field. The centre-to-corner distance of a square is always \(\frac{a}{\sqrt2}\).

Answer 1

The Correct Option is D

Concept: The magnetic field due to a long straight current-carrying conductor at a perpendicular distance \(r\) from it is given by \[ B=\frac{\mu_0 I}{2\pi r}. \] The direction of the magnetic field is determined by the Right-Hand Thumb Rule:
• Thumb in the direction of current.
• Curling fingers give the direction of magnetic field lines. To find the resultant magnetic field at the centre of the square, we first calculate the magnetic field due to each wire and then add them vectorially.

Step 1: Determine the distance of the centre from each corner. The side of the square is \(a\). The diagonal of the square is \[ a\sqrt{2}. \] Therefore, the distance of the centre \(O\) from each corner is half the diagonal: \[ r=\frac{a\sqrt{2}}{2} =\frac{a}{\sqrt{2}}. \]

Step 2: Calculate the magnetic field due to one wire at the centre. Using \[ B=\frac{\mu_0 I}{2\pi r}, \] we get \[ B=\frac{\mu_0 I}{2\pi\left(\frac{a}{\sqrt2}\right)} = \frac{\mu_0 I\sqrt2}{2\pi a}. \] Thus, each wire produces a magnetic field of magnitude \[ B_0=\frac{\mu_0 I\sqrt2}{2\pi a}. \]

Step 3: Determine the directions of the individual magnetic fields. Applying the right-hand thumb rule carefully:
• Wire \(A\) carries current upward.
• Wires \(B\), \(C\) and \(D\) carry current downward. The magnetic field at the centre due to each wire is directed along one of the diagonals. Resolving the fields, it is found that the horizontal and vertical components add in such a way that the resultant magnetic field is directed along the diagonal \(OA\). Each field makes an angle of \(45^\circ\) with the coordinate directions.

Step 4: Add the magnetic field vectors. The component of each field along the diagonal \(OA\) is \[ B_0\cos45^\circ = \frac{\mu_0 I\sqrt2}{2\pi a}\times\frac{1}{\sqrt2} = \frac{\mu_0 I}{2\pi a}. \] All four contributions are along the same direction \(OA\). Therefore, \[ B_{\text{net}} = 4\left(\frac{\mu_0 I}{2\pi a}\right) = \frac{2\mu_0 I}{\pi a}. \] Hence, \[ \boxed{B_{\text{net}}=\frac{2\mu_0 I}{a\pi}} \] and its direction is along \[ \boxed{OA}. \] Therefore, the correct answer is \[ \boxed{\text{(D)}} \]