Four long straight thin wires are held vertically at the corners A, B, Cand D of a square of side ‘a’, kept on a table and carry equal current T.The wire at A carries current in upward direction whereas the current inthe remaining wires flows in downward direction. The net magnetic fieldat the centre of the square will have the magnitude :

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 T.
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 :

Distance from centre $O$O to each corner of the square:$$r=\frac{a}{\sqrt{2}}$$

Magnetic field due to one long straight wire at distance $r$r:$$B=\frac{\mu_0 I}{2\pi r}$$

Substituting $r=\frac{a}{\sqrt{2}}$​:$$B_0=\frac{\mu_0 I}{2\pi (a/\sqrt{2})} =\frac{\mu_0 I\sqrt{2}}{2\pi a} =\frac{\mu_0 I}{\sqrt{2}\pi a}$$​

So each wire produces field $B_0$ at the centre.

Direction of Fields

Using right-hand rule:

• Wire at A (current upward) → field at centre along OB
• Wire at C (current downward) → field at centre along OB
• Wire at B (current downward) → field along OA
• Wire at D (current downward) → field along OC

Resolving components and adding vectorially:$$B_{\text{net}} = 2B_0$$ $$= 2\left(\frac{\mu_0 I}{\sqrt{2}\pi a}\right) = \frac{\sqrt{2}\mu_0 I}{\pi a}$$

Direction: along OB

Final Answer:

$$\boxed{\frac{\mu_0 I\sqrt{2}}{\pi a} \text{ directed along OB}}$$

Correct Answer : Option (c)