Question 58: Magnetism and Matter

Question

A proton and a deutron $(q=+\mathrm{e}, m=2.0 \mathrm{u})$ having same kinetic energies enter a region of uniform magnetic field $\vec{B}$, moving perpendicular to $\vec{B}$. The ratio of the radius $r_d$ of deutron path to the radius $r_p$ of the proton path is:

Accepted Answer

To solve for the ratio of the radii of deutron and proton paths in a magnetic field, we use the formula for the radius $r$ of the circular path of a charged particle moving perpendicular to a uniform magnetic field:
$r = \frac{mv}{qB}$
where:
$m$ is the mass of the particle,
$v$ is the velocity of the particle,
$q$ is the charge of the particle, and
$B$ is the magnetic field strength.
The proton and the deutron are given to have the same kinetic energy. The kinetic energy $K$ of a particle is given by:
$K = \frac{1}{2}mv^2$
From the kinetic energy, we can express the velocity as:
$v = \sqrt{\frac{2K}{m}}$
Since both particles have the same kinetic energy and the charge of the deutron is the same as the charge of the proton $q = e$, but the mass of the deutron is twice that of the proton ($m_d = 2m_p$), substituting the expression for $v$ in the radius formula, we get:
For the deutron:
$r_d = \frac{m_d\sqrt{2K/m_d}}{eB} = \sqrt{\frac{2K}{eB^2}} \cdot \sqrt{m_d}$
For the proton ($m_p = m$):
$r_p = \sqrt{\frac{2K}{eB^2}} \cdot \sqrt{m_p}$
The ratio of the radius of the deutron path $r_d$ to the radius of the proton path $r_p$ is therefore:
$\frac{r_d}{r_p} = \frac{\sqrt{m_d}}{\sqrt{m_p}} = \sqrt{\frac{2m_p}{m_p}} = \sqrt{2}$
So, the correct answer is:
Option C: $\sqrt{2}: 1$.

Answer

$1: 2$

Answer

$1: 1$

Answer

$\sqrt{2}: 1$

Answer

$1: \sqrt{2}$

Magnetism and Matter

A proton and a deutron (q=+\mathrm{e}, m=2.0 \mathrm{u}) having same kinetic energies enter a region of uniform magnetic field \vec{B}, moving perpendicular to \vec{B}. The ratio of the radius r_d of deutron path to the radius r_p of the proton path is:

Choose the correct answer: