The magnitude of magnetic induction at mid-point $$O$$ due to current arrangement as shown in figure will be

Solution :

Magnetic field due to a semi-infinite straight current carrying wire at perpendicular distance \(r\) is :

$$B = \frac{\mu_0 I}{4\pi r}$$

At point \(O\), there are four semi-infinite wires each at distance :

$$r = a$$

Using right hand thumb rule, magnetic field due to each wire at \(O\) is directed into the plane of paper.

Therefore, fields add together.

Magnetic field due to one wire :

$$B_1 = \frac{\mu_0 I}{4\pi a}$$

Total magnetic field :

$$B = 4B_1$$

$$= 4\left(\frac{\mu_0 I}{4\pi a}\right)$$

$$= \frac{\mu_0 I}{\pi a}$$

Final Answer :

$$\frac{\mu_0 I}{\pi a}$$