Question

Two identical circular wires of radius 20 cm and carrying current \(\sqrt2 \)A are placed in perpendicular planes as shown in figure. The net magnetic field at the centre of the circular wires is _________ × 10^{-8} T.
 

Answer 1

Correct Answer: 628

\(\begin{aligned} & \overrightarrow{B_{n e t}}=\frac{\mu_0 i}{2 r} \hat{i}+\frac{\mu_0 i}{2 r} \hat{\jmath} \\ & \Rightarrow B_{\text {net }}=\frac{\mu_0 i}{2 r} \sqrt{2}=4 \pi \times 10^{-7} \times \sqrt{2} \times \sqrt{2} \times \frac{1}{2 \times 0.2}=2 \times 3.14 \times 10^{-6}=628 \times 10^{-8} \mathrm{~T} \end{aligned}\)

Answer 2

Magnetic Field of Perpendicular Circular Loops 

Step 1: Magnetic Field due to a Circular Loop

The magnetic field at the center of a circular loop of radius \( r \) carrying current \( i \) is given by:

\( B = \frac{\mu_0 i}{2r} \)

where \( \mu_0 = 4\pi \times 10^{-7} \text{ T m/A} \) is the permeability of free space.

Step 2: Magnetic Field due to Each Loop

The two loops are identical, with radius \( r = 20 \text{ cm} = 0.2 \text{ m} \) and current \( i = \sqrt{2} \text{ A} \). The magnetic field due to each loop at the center has a magnitude of:

\( B = \frac{\mu_0 i}{2r} = \frac{(4\pi \times 10^{-7} \text{ T m/A}) (\sqrt{2} \text{ A})}{2(0.2 \text{ m})} \)

Step 3: Net Magnetic Field

Since the loops are in perpendicular planes, their magnetic fields at the center are perpendicular to each other. Let the magnetic field due to one loop be along the x-axis and the magnetic field due to the other loop be along the y-axis. The net magnetic field (\( \vec{B}_{net} \)) is the vector sum of the individual fields:

\( \vec{B}_{net} = \frac{\mu_0 i}{2r} \hat{i} + \frac{\mu_0 i}{2r} \hat{j} \)

The magnitude of the net magnetic field is:

\( B_{net} = \sqrt{\left( \frac{\mu_0 i}{2r} \right)^2 + \left( \frac{\mu_0 i}{2r} \right)^2} = \sqrt{2} \frac{\mu_0 i}{2r} = \frac{\mu_0 i \sqrt{2}}{2r} \)

\( B_{net} = \frac{(4\pi \times 10^{-7} \text{ T m/A}) (\sqrt{2} \text{ A}) \sqrt{2}}{2(0.2 \text{ m})} = \frac{4\pi \times 10^{-7} \times 2}{0.4} = 2\pi \times 10^{-6} \text{ T} \)

Using \( \pi = 3.14 \):

\( B_{net} = 2 \times 3.14 \times 10^{-6} \text{ T} = 6.28 \times 10^{-6} \text{ T} = 628 \times 10^{-8} \text{ T} \)

Conclusion:

The net magnetic field at the center is \( \mathbf{628 \times 10^{-8} \text{ T}} \).