Question

The magnitude of magnetic induction at mid point 'O' due to current arrangement as shown in figure will be (\(μ_0\) = permeability of free space)

• A. \(\frac{μ_0I}{2πa}\)
• B. zero
• C. \(\frac{μ_0I}{4πa}\)
• D. \(\frac{μ_0I}{πa}\)

Hint:

Only the two vertical semi-infinite wires matter at O; the horizontal wires point toward O.

Answer 1

The Correct Option is D

Step 1: Read the figure
Current runs along \(AB\), turns down along \(BC\), and along \(DE\) turns up along \(ET\). The vertical wires are separated by \(a\), so each is at distance \(\frac a2\) from \(O\). The horizontal wires lie along the line through \(O\) and give no field there.

Step 2: Field of one semi-infinite wire
The foot of the perpendicular from \(O\) is at the end of each vertical wire, so each gives \(B = \frac{\mu_0I}{4\pi\left(\frac a2\right)} = \frac{\mu_0I}{2\pi a}\).

Step 3: Directions
Wire \(BC\) (current downward, to the left of \(O\)) and wire \(ET\) (current upward, to the right of \(O\)) both give a field out of the page at \(O\) by the right-hand rule. So the fields add.

Step 4: Total
\(B = 2\times\frac{\mu_0I}{2\pi a} = \frac{\mu_0I}{\pi a}\). Option (D).

Final Answer:
The field at O is mu0 I / (pi a). \[ \boxed{\text{(D)}\ \frac{\mu_0I}{\pi a}} \]