To completely demagnetize the bar magnet, the reverse magnetic intensity $H$ produced by the solenoid must equal the coercivity of the magnet:

$$H = 140\,\text{A/m}$$

The magnetic intensity inside an ideal long solenoid is:

$$H = n I = \frac{N}{L} I$$

Given:

• $N = 112$ turns
• $L = 1.6\,\text{m}$
• $H = 140\,\text{A/m}$

The circuit consists of a major circular arc of radius $R$ subtending an angle $\theta = 360^\circ - 90^\circ = 270^\circ = \frac{3\pi}{2}$ radians at $P$, and straight incoming/outgoing leads.

• Magnetic field due to the circular arc subtending $\theta = \frac{3\pi}{2}$:
  $$B_{\text{arc}} = \frac{\mu_0 I}{4\pi R} \theta = \frac{\mu_0 I}{4\pi R} \left(\frac{3\pi}{2}\right)$$
• The straight wire leads contribute in the opposite direction or reduce the field at point $P$ by a factor corresponding to their geometry (giving the term $-2$):
  $$B_{\text{net}} = \frac{\mu_0 I}{4\pi R} \left(\frac{3\pi}{2} - 2\right)$$

The radius of the circular path of a charged particle of mass $m$, charge $q$, and kinetic energy $K$ moving perpendicular to a magnetic field $B$ is:

$$r = \frac{m v}{q B} = \frac{\sqrt{2 m K}}{q B}$$

For a proton ($p$) and an $\alpha$-particle ($\alpha$):

• $m_p = m$, $q_p = e$
• $m_\alpha = 4m$, $q_\alpha = 2e$

Taking the ratio of their radii for identical kinetic energy $K$ and field $B$:

$$\frac{r_p}{r_\alpha} = \frac{\sqrt{m_p}}{q_p} \times \frac{q_\alpha}{\sqrt{m_\alpha}} = \frac{\sqrt{m}}{e} \times \frac{2e}{\sqrt{4m}} = 1 \times \frac{2}{2} = 1$$