Question

The magnetic field at the centre of the circular coil of radius 1 cm and carrying current of 4 A is

• A. $\frac{8\pi}{3}\times10^{-4}T$
• B. $\frac{8\pi}{3}\times10^{-5}T$
• C. $2\pi\times10^{-5}T$
• D. $2\pi\times10^{-4}T$

Answer 1

Answer: (C)

$2\pi\times10^{-5}T$

Answer 2

The problem asks for the magnetic field at the center of a circular coil. However, the figure shows a circular arc, not a full coil. The current 'i' is shown flowing along the arc.

Given values: Radius of the circular arc, $R = 1 \text{ cm} = 1 \times 10^{-2} \text{ m}$. Current flowing through the arc, $I = 4 \text{ A}$.

From the figure, an angle of $90^\circ$ is marked between the radii OA and OB.

The angle explicitly marked in the diagram is $90^\circ$. So, the angle subtended by the current-carrying arc at the center is $\theta = 90^\circ$. In radians, $\theta = 90^\circ \times \frac{\pi}{180^\circ} = \frac{\pi}{2}$ radians.

Using the formula for the magnetic field at the center of a circular arc: $B = \frac{\mu_0 I}{4\pi R} \theta$ Substituting the values: $B = \frac{(4\pi \times 10^{-7} \text{ T m/A}) \times (4 \text{ A})}{4\pi \times (1 \times 10^{-2} \text{ m})} \times \frac{\pi}{2}$ $B = \frac{10^{-7} \times 4}{10^{-2}} \times \frac{\pi}{2}$ $B = 4 \times 10^{-5} \times \frac{\pi}{2}$ $B = 2\pi \times 10^{-5} \text{ T}$

This result matches option (c). The straight segments OA and OB do not contribute to the magnetic field at O because the point O lies on the line of these segments.