Solution

The centripetal force on an electron moving with the speed $v$ perpendicular to the magnetic field $B$ along a circular arc of radius $r$ is

$$\frac{mv^2}{r}=evB\qquad\Rightarrow\qquad B=\frac{p}{er}=\frac{p\sin\alpha}{el}.$$

Nonrelativistically, the momentum $p$ can be determined from the kinetic energy $E$ as $p=\sqrt{2m_0E}=\sqrt{2E_0E}/c$, where $E_0\approx51\times10^4$ eV is the rest energy of an electron. Consequently,

$$B=\frac{\sqrt{2E_0E}\,\sin\alpha}{elc}\approx0.00412\mbox{ T}.$$

Relativistically, the momentum $p$ can be determined from the relation

$$(E_0+E)^2=(pc)^2+E_0^2\qquad\Rightarrow\qquad p=\frac{\sqrt{(2E_0+E)E}}{c}.$$

As a result,

$$B=\frac{\sqrt{(2E_0+E)E}\,\sin\alpha}{elc}\approx0.00416\mbox{ T}.$$

Thus, the nonrelativistic result is about $1$% lower than the relativistic one.