MEC CYU 10

Solution to Question 10

Problem: A single-layer coil bursts due to magnetic pressure.

Physics:
Magnetic field inside solenoid: $B = \mu_0 n I = \mu_0 \frac{1}{d} I$ (since $n=1/d$).
Force on wire per unit length: $f = I \times B_{avg} = I \times (\frac{1}{2} B_{inside}) = \frac{\mu_0 I^2}{2d}$.
Hoop Stress equation: $2 T = f \times (\text{Diameter of loop}) = f \cdot 2R$.
$T = f R = \frac{\mu_0 I^2 R}{2d}$.

Breaking Stress $\sigma_b = \frac{T}{Area} = \frac{T}{\pi d^2 / 4}$.
Substitute $T$:

$$ \sigma_b = \frac{4}{\pi d^2} \cdot \frac{\mu_0 I^2 R}{2d} = \frac{2 \mu_0 I^2 R}{\pi d^3} $$

Solve for $I$:

$$ I = \sqrt{ \frac{\sigma_b \pi d^3}{2 \mu_0 R} } $$