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: Understanding the figure.
The wire has two straight parts and a circular arc. The dashed lines show that the straight wires, if extended, pass through the centre O. The arc covers \(360^{\circ} - 90^{\circ} = 270^{\circ}\), which is \(\frac{3}{4}\) of a circle.

Step 2: Straight parts.
A point that lies on the line of a straight current has \(d\vec{l} \times \vec{r} = 0\). So the two straight wires give zero field at O.

Step 3: Arc part.
A full circle gives \(\frac{\mu_0 I}{2r}\) at the centre. Three quarters of it gives \[ B = \frac{3}{4}\cdot\frac{\mu_0 I}{2r} = \frac{3}{4} \times \frac{4\pi \times 10^{-7} \times 10}{2 \times 0.03} \]

Step 4: Calculate.
\[ B = 0.75 \times 2.094 \times 10^{-4} = 1.57 \times 10^{-4} \text{ T} \]

Step 5: Check the options.
\(1.57 \times 10^{-3}\) T is ten times too large. \(1 \times 10^{-4}\) T and \(2.41 \times 10^{-5}\) T do not match. The right value is option 3.

Final Answer:
The field at O is \(1.57 \times 10^{-4}\) T. \[ \boxed{1.57 \times 10^{-4} \text{ T}} \]