A solenoid having area A and length ' l ' is filled with a material having relative permeability 2. The magnetic energy stored in the solenoid is :

Solution

Physics sir · 4 steps

Energy density in a magnetic material is B²/2μ, so the factor 2 from μᵣ goes into the denominator.

1. Energy density. Inside a medium of permeability μ, the magnetic energy per unit volume is u = B²/2μ. For vacuum, μ = μ₀ gives the familiar B²/2μ₀.
2. Use the material. Here μ = μᵣ μ₀ = 2μ₀, so u = B²/(2 × 2μ₀) = B²/4μ₀.
3. Multiply by volume. The field fills the solenoid's volume V = Al, so the total energy is u × V.

   U = B²/2μᵣμ₀ × Al = (B² A l)/4μ₀

4. So. The stored energy is (B² A l)/4μ₀, which is option (a).

The idea to keep. Magnetic energy density is B²/2μ with μ = μᵣμ₀. For a given B, a larger μᵣ means less energy stored, so μᵣ always divides.

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