Problem 28 - Master Resource Book in JEE Main Physics

Question

A deutron of kinetic energy \(50 \mathrm{keV}\) is describing a circular orbit of radius \(0.5 \mathrm{~m}\), is plane perpendicular to magnetic field \(B\). The kinetic energy of proton that describes a circular orbit of radius \(0.5 \mathrm{~m}\) in the same plane with the same magnetic field \(B\), is (a) \(200 \mathrm{keV}\) (b) \(50 \mathrm{keV}\) (c) \(100 \mathrm{keV}\) (d) \(25 \mathrm{keV}\)

Step-by-Step Solution

Answer

The kinetic energy of the proton is 100 keV.

Step 1: Understand the Problem

We need to find the kinetic energy of a proton in a magnetic field where a deuteron with known kinetic energy and radius is moving. Note that a deuteron is a nucleus of deuterium, consisting of a proton and a neutron.

Step 2: Use the Magnetic Force Equation

Consider the equation for the magnetic force providing the centripetal force: \( qvB = \frac{mv^2}{r} \). From this, we can derive the velocity \( v = \frac{qBr}{m} \).

Step 3: Relate Kinetic Energy to Velocity

The kinetic energy \( KE \) is given by \( KE = \frac{1}{2}mv^2 \). Substitute the value of velocity derived earlier into this kinetic energy formula: \( KE = \frac{1}{2}m\left(\frac{qBr}{m}\right)^2 \).

Step 4: Compare Proton and Deuteron

The charge \( q \) is the same for both proton and deuteron, but the mass of a deuteron is approximately twice the mass of a proton (\( m_d = 2m_p \)). This implies that the kinetic energy \( KE \) for the proton should be compared given \( m_d = 2m_p \).

Step 5: Kinetic Energy Ratio

Using the derived formula for kinetic energy, note that \( KE \propto \frac{1}{m} \). Therefore, the proton, which is half the mass of the deuteron, will have twice the kinetic energy. If the deuteron has 50 keV, the proton will have 100 keV.

Key Concepts

Kinetic Energy

Kinetic energy is a measure of the energy that a particle possesses due to its motion. It depends on both the mass of the particle and the square of its velocity. The formula for kinetic energy is given as:\[ \text{KE} = \frac{1}{2}mv^2 \]where \( m \) represents the mass, and \( v \) is the velocity of the particle.
For particles moving in a magnetic field, their velocity can be influenced by the field's strength and the nature of their circular path. In this scenario, the velocity is derived from balancing the magnetic force and the centripetal force needed for circular motion. This interplay demonstrates the underlying principles of energy transformations and force balances in physics.
When comparing different particles, such as a proton and a deuteron, their masses play a crucial role. The deuteron is roughly twice the mass of a proton, hence for the same circular path radius in a magnetic field, the kinetic energy of the proton will be calculated to be double that of the deuteron, given they move in similar conditions.

Circular Motion in Magnetic Fields

Circular motion in magnetic fields is governed by the interaction between the charged particle and the magnetic field. The magnetic force acts as the centripetal force, keeping the particle moving in a circular path. This relationship is expressed as:\[ qvB = \frac{mv^2}{r} \]where \( q \) is the charge, \( v \) is the velocity, \( B \) is the magnetic field strength, and \( r \) is the radius of the circular path.
To find the velocity of a particle moving in a circular path inside a magnetic field, we rearrange the equation to get \( v = \frac{qBr}{m} \). This shows that the velocity is directly proportional to the radius \( r \) and the magnetic field strength \( B \), and inversely proportional to the mass \( m \).
Such equations allow us to understand how particles like protons and deuterons behave when subjected to magnetic fields. By examining their behavior, we can derive insights about their energy levels and motion patterns, leading to applications in particle accelerators and other technology relying on magnetic fields.

Proton and Deuteron Dynamics

Understanding the dynamics of protons and deuterons in magnetic fields involves considering their mass and charge differences. While both particles carry the same type of charge, their masses are different, with the deuteron being twice as massive as the proton.
This mass difference impacts how each particle reacts to the same magnetic field. As noted earlier, the kinetic energy of a particle in a circular path is inversely related to its mass for a constant radius and magnetic field. Hence, a proton, having half the mass of a deuteron, achieves double the kinetic energy while maintaining the same circular path\(50 \mathrm{~m}\) and conditions.
Thus, for the proton moving in the same magnetic field as the deuteron, the calculated kinetic energy would be \(100 \mathrm{keV}\), given the deuteron's kinetic energy is \(50 \mathrm{keV}\). This simple yet fundamental principle illustrates how mass impacts energy dynamics in magnetic fields, serving as a basis for numerous applications in physics.