Question

A wire shown in figure carries a current of 10 A. The magnitude of the magnetic field at the centre O is: (Given: radius of the bent coil is 3 cm.)

• A. \(1 \times 10^{-4}\) T
• B. \(1.57 \times 10^{-3}\) T
• C. \(1.57 \times 10^{-4}\) T
• D. \(2.41 \times 10^{-5}\) T

Hint:

Straight wires through O give zero. The arc is 270 degrees, so use three quarters of \(\mu_0 I/2r\).

Answer 1

The Correct Option is C

Step 1: Split the wire.
Use the Biot-Savart law on each part separately and add the results.

Step 2: Straight pieces.
Each straight piece lies along a line through O, so the angle between \(d\vec l\) and \(\vec r\) is \(0\) or \(\pi\). Then \(\sin\theta = 0\) and these pieces add nothing.

Step 3: Arc piece.
For the arc, \(dB = \dfrac{\mu_0 I\, dl}{4\pi r^2}\). Add over arc length \(l = r\theta\) with \(\theta = \dfrac{3\pi}{2}\): \[ B = \frac{\mu_0 I \theta}{4\pi r} = \frac{10^{-7}\times 10 \times 4.712}{0.03} \]

Step 4: Number.
\[ B = \frac{4.712\times10^{-6}}{0.03} = 1.57\times10^{-4} \text{ T} \]

Step 5: Answer.
This matches option 3.

Final Answer:
The field is \(1.57\times10^{-4}\) T, option 3. \[ \boxed{1.57 \times 10^{-4} \text{ T}} \]