Question

A proton carrying 1 MeV kinetic energy is moving in a circular path of radius R in uniform magnetic field. What should be the energy of an $\alpha$-particle to describe a circle of same radius in the same field?

• A. 2 MeV
• B. 1 MeV
• C. 0.5 MeV
• D. 4 MeV

Answer 1

The Correct Option is B

To solve this problem, we need to understand the relationship between the radius of the circular path, the kinetic energy of the charged particle, and the magnetic field.

When a charged particle moves in a magnetic field, it experiences a centripetal force that makes it move in a circular path. The centripetal force is provided by the magnetic Lorentz force. Therefore, for a proton moving in a circular path:

F_{\text{centripetal}} = F_{\text{magnetic}}

The expression for centripetal force is:

F_{\text{centripetal}} = \frac{mv^2}{R}

where:

• m is the mass of the particle,
• v is the velocity of the particle,
• R is the radius of the path.

The magnetic force is given by:

F_{\text{magnetic}} = qvB

where:

• q is the charge of the particle,
• v is the velocity of the particle,
• B is the magnetic field strength.

Setting F_{\text{centripetal}} = F_{\text{magnetic}} gives:

\frac{mv^2}{R} = qvB

Simplifying this expression for velocity, v:

v = \frac{qBR}{m}

The kinetic energy (K.E) of the particle is related to its velocity by:

\text{K.E} = \frac{1}{2}mv^2

Substituting v = \frac{qBR}{m} in the kinetic energy:

\text{K.E} = \frac{1}{2}m\left(\frac{qBR}{m}\right)^2 = \frac{q^2B^2R^2}{2m}

Let's compare the kinetic energies of the proton and the alpha particle. For the same radius (R) and magnetic field (B), the equation becomes:

\frac{q^2}{m}_{\text{proton}} = \frac{q^2}{m}_{\alpha\text{-particle}}

For a proton: charge (q_p = e), mass (m_p = m)
For an alpha particle: charge (q_{\alpha} = 2e), mass (m_{\alpha} = 4m).

Simplifying it, the kinetic energy of the alpha particle:

\text{K.E}_{\alpha} = \frac{(2e)^2B^2R^2}{2 \cdot 4m} = \text{K.E}_p

Now, solving for \text{K.E}_{\alpha}:

\text{K.E}_{\alpha} = \text{K.E}_{\text{proton}} = 1 \text{ MeV}

Thus, the energy of the alpha particle needed to describe a circle of the same radius in the same field is 1 MeV.