Two identical circular wires of radius 20 cm and carrying current √2 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.
(Take π=3.14)

Solution

Physics sir · 4 steps

Each loop gives the same field at the centre, but at right angles, so add them like vectors.

1. One loop. At the centre of a circular loop, Biot-Savart gives B = (μ₀ I)/2R. The field points along the loop's axis, perpendicular to its plane.

   B₁ = B₂ = (4π × 10⁻⁷ × √2)/(2 × 0.2) = √2 π × 10⁻⁶ T

2. Directions. The horizontal loop gives a field along the Y-axis. The vertical loop gives a field along the X-axis. These two fields are perpendicular to each other.
3. Add as vectors. Bₙₑₜ = √(B₁² + B₂²) = √2 B₁ = √2 × √2 π × 10⁻⁶ = 2π × 10⁻⁶ T.
4. So. With π = 3.14, Bₙₑₜ = 6.28 × 10⁻⁶ T = 628 × 10⁻⁸ T. The answer is 628.

The common slip: 444. Stopping after one loop gives √2 π × 10⁻⁶ T, about 444, not 628. The second, perpendicular loop multiplies the field by another √2.

The idea to keep. The field at a loop's centre lies along its axis. For loops in different planes, find each field's direction first, then add them as vectors, never as plain numbers.

Answer from the NTA final answer key. Solution by PrepWiser.