Problem 1909 - JEE Physics

Question

A 2 Mev proton is moving perpendicular to a uniform magnetic field of \(2.5\) tesla. The force on the proton is (a) \(3 \times 10^{-10} \mathrm{~N}\) (b) \(70.8 \times 10^{-11} \mathrm{~N}\) (c) \(3 \times 10^{-11} \mathrm{~N}\) (d) \(7.68 \times 10^{-12} \mathrm{~N}\)

Step-by-Step Solution

Answer

The force on the proton is approximately \(4.34 \times 10^{-12}\,\text{N}\). However, none of the given options match this result, which may indicate an error in the options provided.

Step 1: Convert energy to Joules

Given the energy of the proton is 2 MeV (Mega electron volts), we will convert it to Joules using the following conversion factor: 1 eV = \(1.6 \times 10^{-19}\) J 1 MeV = \(10^6\) eV So, \(2\,\text{MeV} = 2 \times 10^6 \times 1.6 \times 10^{-19}\, \text{J} = 3.2 \times 10^{-13}\,\text{J}\)

Step 2: Calculate the velocity of the proton

To calculate the velocity of the proton, we need to consider the kinetic energy formula, \(E = \frac{1}{2}mv^2\), where \(E\) is the kinetic energy, \(m\) is the mass of the proton, and \(v\) is the velocity of the proton. The mass of a proton, \(m = 1.67 \times 10^{-27}\,\text{kg}\). Rearrange the formula to find the velocity: \(v = \sqrt{\frac{2E}{m}}\) Substitute the values: \(v = \sqrt{\frac{2 \times 3.2 \times 10^{-13}\text{J}}{1.67 \times 10^{-27}\,\text{kg}}}\) Calculating the velocity, we get: \(v = 1.082 \times 10^{7}\, \text{m/s}\)

Step 3: Calculate the force on the proton

Now, we can calculate the force on the proton using the Lorentz force formula: \(F = qvB\). The charge of a proton, \(q = 1.6 \times 10^{-19}\,\text{C}\), and the magnetic field strength, \(B = 2.5\,\text{T}\). Substitute the values: \(F = (1.6 \times 10^{-19}\text{C}) \times (1.082 \times 10^7\,\text{m/s}) \times (2.5\, \text{T})\) Calculating the force, we get: \(F \approx 4.34 \times 10^{-12}\,\text{N}\) Comparing our answer to the given options, it seems none of them match our result, which might indicate that there might be a mistake in the options given in the exercise.

Key Concepts

Kinetic Energy Conversion

In physics, converting energy between different units is crucial for calculations. Here, we deal with MeV (Mega electron volts) and Joules. When working with particle energies, such as protons, MeV is a common unit. To align these energy levels with everyday physics problems, we often convert them into Joules, as this unit allows us to apply conventional physics equations more easily. For example:

• 1 electron volt (eV) = \(1.6 \times 10^{-19}\) Joules
• 1 Mega electron volt (MeV) = \(10^6\) eV = \(1.6 \times 10^{-13}\) Joules

In this problem, a proton has 2 MeV. Converting this to Joules, we use:\[ 2 \times 10^6 \times 1.6 \times 10^{-19} = 3.2 \times 10^{-13}\text{ J} \] This energy conversion sets the stage for further calculations involving kinetic energy and velocity.

Lorentz Force

Lorentz force is the force exerted on a charged particle moving through a magnetic field. When a proton (a charged particle) moves perpendicular to a magnetic field, the force is determined by:\[ F = qvB \]Here,

• \( F \) is the Lorentz force
• \( q \) is the charge of the proton, \(1.6 \times 10^{-19} \,\text{Coulombs}\)
• \( v \) is the velocity of the proton
• \( B \) is the magnetic field strength, given as 2.5 Tesla in this problem

The motion of the proton causes it to experience this force, allowing us to calculate the exact impact of the magnetic field on its trajectory. Understanding the Lorentz force is crucial when dealing with electromagnetic fields and charged particles.

Velocity Calculation

After converting kinetic energy into Joules, we determine the velocity of a proton using the kinetic energy formula:\[ E = \frac{1}{2}mv^2 \]Rearranged for velocity, this becomes:\[ v = \sqrt{\frac{2E}{m}} \]Here:

• \( E \) is the kinetic energy, found to be \(3.2 \times 10^{-13}\text{ J}\)
• \( m \) is the mass of the proton, approximately \(1.67 \times 10^{-27}\text{ kg}\)

Plugging these values into our rearranged formula gives:\[ v = \sqrt{\frac{2 \times 3.2 \times 10^{-13} \text{ J}}{1.67 \times 10^{-27} \text{ kg}}} \approx 1.082 \times 10^{7}\text{ m/s} \]Understanding the speed of protons in a magnetic field helps in planning experiments and predicting behavior in various conditions.

Proton Charge

The proton charge is a fundamental constant in physics:- Every proton carries a positive charge of \(1.6 \times 10^{-19}\text{ Coulombs}\).
This charge is critically important for calculations involving electric and magnetic fields. A proton's motion in these fields leads to the generation of forces like the Lorentz force.When calculating the interaction of particles with fields, always remember:

• The charge determines the direction and magnitude of the force
• Protons, due to this charge, are influenced in a compound manner when traveling across magnetic or electric fields

This is foundational for understanding the behavior of particles within electromagnetic fields and is applied in the calculation of forces acting on particles, thereby affecting how they move and interact in fields.