The value of aluminium susceptibility is 2.2 × 10⁻⁵. The percentage increase in the magnetic field if space within a current carrying toroid is filled with aluminium is x/10⁴. Then the value of x is _________________.

Solution

Physics sir · 4 steps

Filling the toroid multiplies B by (1+χ), so the fractional increase is just χ itself.

1. Empty toroid. With air inside, the field is B₀ = μ₀ n I, where n is the number of turns per unit length.
2. Filled with aluminium. The material adds its own magnetisation M = χ H, so B = μ₀(1+χ)nI = (1+χ)B₀. The change is Δ B = χ B₀.

   (Δ B)/B₀× 100 = χ × 100 = 2.2 × 10⁻⁵ × 100 = 2.2 × 10⁻³ %

3. Match the form. 2.2 × 10⁻³ = 22/10⁴, so the percentage increase is 22/10⁴.
4. So. x = 22.

The common slip: 0.22. Stopping at the fraction χ = 2.2 × 10⁻⁵ and forgetting to multiply by 100 for a percentage gives 0.22, not 22.

The idea to keep. For any material filling a solenoid or toroid, B = (1+χ)B₀. The fractional change is χ, and the percentage change is 100χ.

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