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

Concept: Magnetic field due to a long straight current-carrying wire: \[ B = \frac{\mu_0 I}{2\pi r} \] Key ideas:

Distance from centre to each corner of square: \( r = \frac{a}{\sqrt{2}} \)
Direction of magnetic field determined by right-hand thumb rule
Vector addition of magnetic fields

Step 1: Magnetic field magnitude due to each wire. Distance from centre to each corner: \[ r = \frac{a}{\sqrt{2}} \] Thus field due to each wire: \[ B_0 = \frac{\mu_0 I}{2\pi r} = \frac{\mu_0 I}{2\pi \left(\frac{a}{\sqrt{2}}\right)} = \frac{\mu_0 I \sqrt{2}}{2\pi a} \]
Step 2: Directions using right-hand rule. - Wire at A: current upward → field direction anticlockwise. - Wires at B, C, D: current downward → field clockwise. At the centre, magnetic fields are tangential to circles around wires. Resolve each field into components along diagonals.
Step 3: Symmetry analysis. Due to square symmetry:

Fields from B and D cancel partially along one diagonal.
Field from C adds with resultant of others.
Net field lies along diagonal OB.

Step 4: Resultant magnitude. Each magnetic field makes \( 45^\circ \) with diagonals. Effective components add vectorially, giving: \[ B_{\text{net}} = 2B_0 \] \[ B_{\text{net}} = 2 \times \frac{\mu_0 I \sqrt{2}}{2\pi a} = \frac{\mu_0 I \sqrt{2}}{\pi a} \]
Step 5: Direction. From vector addition, resultant is along diagonal OB.

Answer 2

Instead of resolving all four fields into components individually, this pairing method groups the wires by their position relative to the centre and shows that two of them cancel completely, leaving only two wires to determine the answer.

Distance from centre to each wire: The centre O of the square is equidistant from every corner, at a distance \( r = \dfrac{a}{\sqrt{2}} \) (half the diagonal). So each wire produces a field of the same magnitude at O:

\[ B_0 = \frac{\mu_0 I}{2\pi r} = \frac{\mu_0 I}{2\pi \left(\frac{a}{\sqrt{2}}\right)} = \frac{\mu_0 I \sqrt{2}}{2\pi a} \]

Pairing the diagonally opposite wires B and D: B and D sit on opposite ends of the same diagonal, at equal distance from O, and both carry current in the same (downward) direction. Because a straight wire's field circles it in a fixed sense, two identical wires placed symmetrically on either side of a point, carrying current in the same direction, produce fields at that point that point in exactly opposite directions. So the contributions of B and D at O cancel each other out completely.

What's left, wires A and C: A and C also sit on opposite ends of a diagonal, but they carry current in opposite directions (A upward, C downward). This reversal of current direction, combined with their opposite positions, means their individual field directions at O actually line up rather than opposing, so their contributions reinforce each other instead of cancelling.

Net field: With B and D cancelling, the net field is just twice the field due to one wire (A or C alone):

\[ B_{\text{net}} = 2B_0 = 2 \times \frac{\mu_0 I \sqrt{2}}{2\pi a} = \frac{\mu_0 I \sqrt{2}}{\pi a} \]

Since the reinforcing contributions come from wires A and C acting along the diagonal perpendicular to AC, the resultant field is directed along OB.

So, the net magnetic field at the centre has magnitude \( \dfrac{\mu_0 I \sqrt{2}}{\pi a} \), directed along OB.