(\mathrm{KE})_\alpha = 1:4$

Magnetic field strength: $B = 3 \, \mathrm{T}$

Angle with the magnetic field: $\theta = 90^\circ$

Definitions:

$q$ is the charge of the particle.

$v$ is the velocity of the particle.

$F$ is the magnetic force acting on the particle.

Using the Lorentz Force Equation:

The magnetic force ($F$) on a charged particle moving in a magnetic field is given by:

$ F = qvB \sin \theta $

Since $\theta = 90^\circ$, $\sin 90^\circ = 1$, so:

$ F = qvB $

Given that the ratio of kinetic energy (KE) is:

$ \frac{(\mathrm{KE})_p}{(\mathrm{KE})_\alpha} = \frac{1}{4} $

The kinetic energy ($\mathrm{KE}$) is defined as:

$ \mathrm{KE} = \frac{1}{2} m v^2 $

Thus, we have:

$ \frac{\frac{1}{2} m_p v_p^2}{\frac{1}{2} m_\alpha v_\alpha^2} = \frac{1}{4} $

Since the mass of the alpha particle $(m_\alpha)$ is four times that of the proton $(m_p)$, i.e., $m_\alpha = 4m_p$, the equation simplifies to:

$ \frac{v_p}{v_\alpha} = 1 \implies v_p = v_\alpha $

Calculating the Ratio of Forces:

From the equation $F = qvB$, the forces on the proton and alpha particle are:

For the proton: $F_p = q_p v_p B$

For the alpha particle: $F_\alpha = q_\alpha v_\alpha B$

The charge of the proton is $q_p = e$ and the charge of the alpha particle is $q_\alpha = 2e$. Therefore, we find the ratio of the forces:

$ \frac{F_p}{F_\alpha} = \frac{q_p v_p B}{q_\alpha v_\alpha B} = \frac{e}{2e} = \frac{1}{2} $

Thus, the ratio of the magnetic force acting on the proton to that on the alpha particle is $1:2$.

Moving Charges And Magnetism

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