Generally, the best starting point of  any kinematics calculation for any reaction is from the CM frame. In the exit channel, the four-momentum in the CM frame is

$P_{cm} = \begin{pmatrix} \sqrt{m^2 + p^2} = E_{cm} \\ \vec{p} \end{pmatrix}$

the boost from CM frame to Lab frame is $\vec{\beta}$. Lets define the perpendicular vector on the plane of $\vec{p}$ be $\hat{n}$, thus, the four-momentum in the Lab frame is

$P = \begin{pmatrix} E \\ \vec{k} \end{pmatrix} =\begin{pmatrix} \gamma E_{cm} + \gamma \beta \hat{\beta}\cdot \vec{p} \\ \left( \gamma \beta E_{cm} + \gamma (\hat{\beta} \cdot \vec{p}) \right) \hat{\beta} + (\hat{n} \cdot \vec{p} ) \hat{n} \end{pmatrix}$

where

$\displaystyle E_{cm} = \frac{1}{2E_t}\left( E_t^2 + m_b^2 -m_B^2 \right)$

$\displaystyle p^2 = \frac{1}{4E_t^2} \left(E_t^2 - (m_b + m_B)^2\right) \left(E_t^2 - (m_b - m_B)^2\right)$

$\displaystyle E_t^2 = (m_a+m_A)^2 + 2m_aT$

$\displaystyle \beta = \frac{\sqrt{(m_A+T)^2 - m_A^2}}{m_a + m_A + T}, \gamma = \frac{1}{\sqrt{1-\beta^2}}$

We can see, the particle-A is moving with kinematic energy of $T$, hit particle-a, result in particle-b and particle-B.

Suppose the magnetic field is parallel to z-axis

$\vec{B} = B \hat{z}$

$\displaystyle \rho = \frac{\vec{k} \cdot \hat{xy} }{cZB}$

where $Z$ is the charge state of the particle, and $\hat{xy}$ is the perpendicular unit vector on xy-plane.

The time for 1 cycle is

$\displaystyle T_{cyc} = \frac{2\pi \rho}{v_{xy}} = \frac{2\pi}{cZB} \frac{\vec{k}\cdot\hat{xy}}{v_{xy}} = \frac{2\pi}{c^2 ZB} E$

And the length for 1 cycle is

$\displaystyle z_{cyc} = v_z T_{cyc} = \frac{2\pi}{cZB} \vec{k}\cdot \hat{xy} \frac{v_z}{v_{xy}} = \frac{2\pi}{cZB} \vec{k}\cdot \hat{z}$

When the Lorentz boost is parallel to the B-field direction,

$\vec{\beta} \cdot \hat{z} = 1, \hat{n} \cdot \hat{z} = 0$

$\displaystyle z_{cyc} = \frac{2\pi}{cZB} \left(\gamma \beta E_{cm} + \gamma (\hat{\beta} \cdot \vec{p})\right)$ ………………. (1)

and the total energy is

$\displaystyle E = \gamma E_{cm} + \gamma \beta (\hat{\beta}\cdot \vec{p})$ …………….. (2)

By eliminating the $\hat{\beta} \cdot \vec{p}$,

$\displaystyle E = \frac{1}{\gamma}E_{cm} + \frac{cZB}{2\pi} \beta z_{cyc}$ …………….. (3)

We can see that, the energy and $z_{cyc}$ have a linear relation. That simplified a alot of thing. For instance, the intercept is related to $E_{cm}$, which is further related to the Q-value of the reaction. If we can plot the E-z plot and determine the intercept, we can extract the Q-value, which is related to the excitation energy or some sort.

Without the magnetic field, to extract the Q-value, we really need to know the scattering angle (the azimuthal angle can be skipped due to symmetry in some cases). And usually, the relation is not linear.