Question 44: 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: