Find the mutual inductance in the arrangement, when a small circular loop of wire of radius '' is placed inside a large square loop of wire of side . The loops are coplanar and their centres coincide :
- A.
- B.
- C.
- D.
Show answer and solution
Correct option: D
Solution
Since L ≫ R, the small loop sits at the centre of the square loop where the field is approximately uniform. The magnetic field at the centre of a square loop of side L carrying current I is found by summing contributions from all 4 sides. Each side is at a perpendicular distance L/2 from the centre, and subtends angles of ±45°. The field from one side:
B₁ = (μ₀I)/(4π · L/2) · 2sin45° = (μ₀I)/(4π · L/2) · √2 = (μ₀I√2)/(2πL)
For all 4 sides:
B = 4 · (μ₀I√2)/(2πL) = (2√2 μ₀I)/(πL)
The flux through the small circular loop (area = πR²):
Φ = B · πR² = (2√2 μ₀I)/(πL) · πR² = 2√2 μ₀IR²/L
Therefore, mutual inductance M = Φ/I = 2√2 μ₀R²/L, which is option 4.
B₁ = (μ₀I)/(4π · L/2) · 2sin45° = (μ₀I)/(4π · L/2) · √2 = (μ₀I√2)/(2πL)
For all 4 sides:
B = 4 · (μ₀I√2)/(2πL) = (2√2 μ₀I)/(πL)
The flux through the small circular loop (area = πR²):
Φ = B · πR² = (2√2 μ₀I)/(πL) · πR² = 2√2 μ₀IR²/L
Therefore, mutual inductance M = Φ/I = 2√2 μ₀R²/L, which is option 4.
[JEE Main 2023 · 29 Jan · Shift 1]