Question

ratio of energy density of magnetic field at centre of current carrying loop to that at a distance R/√2 from centre of loop on its axis is​

Answer 1

The ratio of energy density of magnetic field at center of current carrying loop to that at a distance is √3/2 R

Explanation:

The energy density of magnetic field at center is given by the formula:

Bc = (μ₀i)/(2R)

The energy density of magnetic field at a distance from center of loop is given by the formula:

Bx = (μ₀ir)/2(x² + R²)³⁾²

Now, the ratio is given as:

Bc/Bx = ((μ₀i)/(2R))/((μ₀iR)/2(x² + R²)³⁾²)

Bc/Bx = (μ₀i)/(2R) × (2(x² + R²)³⁾²)/((μ₀iR))

Where, x = R/√2

Bc/Bx = (μ₀i)/(2R) × (2((R/√2)² + R²)³⁾²)/((μ₀iR))

Bc/Bx = 1/(2R) × (2(R²/2 + R²)³⁾²)/(R)

Bc/Bx = (2(R²/2 + R²)³⁾²)/2R²

Bc/Bx = (R² + 2R²)³⁾²/2R²

Bc/Bx = (3R²)³⁾²/2R²

On squaring both sides, we get,

(Bc/Bx)² = (3R²)³/4R⁴

(Bc/Bx)² = (3R⁶)/4R⁴

(Bc/Bx)² = 3R²/4

Taking square root on both sides, we get,

∴ Bc/Bx = √3/2 R