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.