Past exam of the short options paper shared by Prelims and Parts A and B of the physics course of the University of Oxford / 2017 / S16 / 3

Use a steady planar sonic ablation model with absorbed intensity $I_a$, critical mass density ${\rho }_c=(A_i/Z)m_u{n}_c$, and $n_c={ϵ}_0{m}_e(2\pi c/\lambda {)}^2/e^2$. In the usual isothermal-electron sonic balance, $I_a\simeq 4{\rho }_c{c}_s^3$ and the ablation pressure is $P_A\simeq 2{\rho }_c{c}_s^2$. Eliminating $c_s$ givesFor $A_i/Z\simeq 2$ and complete absorption, the supplied laser parameters give about $1.15\times 10^{13}$ Pa, or $115$ Mbar. Mach one alone fixes neither the material mass per electron nor the enthalpy/absorption coefficient. Different sonic-flow closures change the order-one constant; incomplete absorption multiplies the estimate by its fraction to power $2/3$. The robust result is the two-thirds scaling, rather than a material-independent numerical pressure.

An inertially confined sphere disassembles on approximately $\tau \sim r/c_s$, while $n_i=\rho /{\bar{m}}_i$. Substitute these in the given number-density confinement product to getFor an equal D-T mix, ${\bar{m}}_i\simeq 2.5m_u$. Taking $c_s\sim \sqrt{2k_{B}T/{\bar{m}}_i}$ at 10 keV gives $A\simeq 0.365$ kg m${}^{-2}=0.0365$ g cm${}^{-2}$. An adiabatic sound-speed convention changes this by an order-one factor. This estimate implements the supplied Lawson product and is not the separate, more stringent alpha-stopping areal-density criterion.

The final radius is $r_f=0.5\mathrm{mm}/15=33.3\mu$m. For a 3.5 MeV alpha particle, $p\simeq \sqrt{2m_{\alpha }{E}_{\alpha }}$ and the largest perpendicular gyroradius is $p/(2e{B}_f)$. Setting it equal to the core radius givesThis is an order-of-magnitude onset estimate for magnetic deflection, assuming flux freezing and a suitably oriented field. A small gyroradius alone does not ensure complete confinement along open field lines or efficient deposition for every pitch angle.

Magnetic pressure is the field energy density $P_B=B_f^2/(2{\mu }_0)$. Using the preceding final field gives $\boxed{P_B\simeq 2.6\times 10^{13}\mathrm{Pa}\simeq 260\mathrm{Mbar}}$. This uses the compressed field, not the initial 36 T field; substituting the latter would omit the square of the already squared convergence amplification.