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

### Related Formula 1. Magnetic Energy Density (energy per unit volume): u_m = fracB^22 mu = fracB^22 mu_r mu_0$$u_m = \frac{B^2}{2 \mu} = \frac{B^2}{2 \mu_r \mu_0}$$ 2. Total Energy stored: U = u_m times V = u_m times (A l)$$U = u_m \times V = u_m \times (A l)$$ ### Core Logic We are given: - Relative permeability mu_r = 2$\mu_r = 2$ - Permeability of the filled core medium mu = mu_r mu_0 = 2 mu_0$\mu = \mu_r \mu_0 = 2 \mu_0$ Substitute mu_r = 2$\mu_r = 2$ into the energy density expression: u_m = fracB^22 (2 mu_0) = fracB^24 mu_0$$u_m = \frac{B^2}{2 (2 \mu_0)} = \frac{B^2}{4 \mu_0}$$ Multiply by the total volume of the solenoid (V = A l$V = A l$): U = u_m cdot V = fracB^24 mu_0 A l$$U = u_m \cdot V = \frac{B^2}{4 \mu_0} A l$$ ### Pattern Recognition Sees: Magnetic energy stored in a solenoid with medium relative permeability. Trap: Placing the relative permeability mu_r$\mu_r$ in the numerator of the formula instead of the denominator. Shortcut: Magnetic energy density is always inversely proportional to permeability. With medium mu = 2mu_0$\mu = 2\mu_0$, energy density is halved compared to free space, giving fracB^24mu_0$\frac{B^2}{4\mu_0}$. Multiply by volume Al$Al$ to get fracB^2 Al4mu_0$\frac{B^2 Al}{4\mu_0}$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Electromagnetic Induction Class 12 Physics: Magnetism and Matter