Solutions

1)

Using Eq. 5.35 in the textbook, where $\theta_2 = -\theta_1 =\pi/4$, and adding the four sides, $B = \sqrt{2} \mu_0 I /\pi a$.

From the symmetry of the problem, $A=0$.

2)

Using $\nabla\times {\bf H} = {\bf J}_f$ (static case), $\oint_C {\bf H}\cdot d{\bf l} = I$, where $C$ is an Amperean loop with a radius $s$ around the axis of the cylinder. From this, we obtain $H = I/2\pi s$.

The magnetic fields are

$${\bf B}({\bf r}) = {\mu_0 \over 4\pi}I\int {d{\bf l}\times ({\bf r -r^\prime}) \over \vert{\bf r -r^\prime}\vert^3 } ~,$$ (14)

$${\bf A}({\bf r}) = {\mu_0 \over 4\pi}\int {{\bf J}~d\tau^\prime \over \vert{\bf r -r^\prime}\vert } ~.$$ (15)

$$B(s) = {\mu_1I \over 2\pi s}~~ (s<R),$$ (16)

$$= {\mu_2I \over 2\pi s}~~ (s>R).$$ (17)

The surface bound current is linearly related to the discontinuity in $B$ parallel to the surface at $s=R$: $\Delta B = \mu_0 K = (\mu_2-\mu_1)I/2\pi R$. So $K=(\mu_2-\mu_1)I/2\pi \mu_0 R$.

3)

The magnetic field at the current loop: $B(t) \approx \mu_0 I(t)/2\pi x$. The flux: $\Phi(t) \approx AB(t) =MI(t)$. The mutual inductance is then $M = \mu_0 A\over 2\pi x$.

The emf: ${\cal E} = -d\Phi/dt = RI_{\rm loop}$. So

$$I_{\rm loop} = -{\mu_0 A\over 2\pi Rx} {dI(t) \over dt}~.$$ (18)

4)

In the centre of mass frame of the collision, the initial and the final total momenta must be zero. Therefore, two photons must be produced in order to conserve momentum. Since the law of physics is invariant in any reference frame, two photons are produced in any reference frame.