Problem 4 - Master Resource Book in JEE Main Physics

Question

A proton of velocity \((3 \hat{\mathbf{i}}+2 \hat{\mathbf{j}}) \times 10^{5} \mathrm{~ms}^{-1}\) enters a magnetic field \((2 \hat{\mathbf{i}}+3 \hat{\mathbf{k}}) \mathrm{T}\). If the specific charge is \(9.6 \times 10^{7} \mathrm{C} \mathrm{kg}^{-1}\), the acceleration of the proton in \(\mathrm{ms}^{-2}\) is (a) \((6 \hat{i}-9 \hat{j}+4 \hat{k}) \times 9.6 \times 10^{12}\) (b) \((6 \hat{\mathrm{i}}+9 \hat{\mathrm{j}}+4 \hat{\mathrm{k}}) \times 9.6 \times 10^{12}\) (c) \((6 \hat{i}-9 \hat{j}-4 \hat{k}) \times 9.6 \times 10^{12}\) (d) \((6 \hat{i}+9 \hat{j}-4 \hat{k}) \times 9.6 \times 10^{12}\)

Step-by-Step Solution

Answer

The correct answer is (c).

Step 1: Use the Lorentz Force Formula

The Lorentz force formula for a charged particle moving in a magnetic field is given by \( \mathbf{F} = q (\mathbf{v} \times \mathbf{B}) \), where \( q \) is the charge, \( \mathbf{v} \) is the velocity vector, and \( \mathbf{B} \) is the magnetic field vector.

Step 2: Determine the Cross Product \( \mathbf{v} \times \mathbf{B} \)

Compute the cross product of the velocity vector \( (3 \hat{\mathbf{i}} + 2 \hat{\mathbf{j}}) \times 10^{5} \mathrm{~ms}^{-1} \) and the magnetic field vector \( (2 \hat{\mathbf{i}} + 3 \hat{\mathbf{k}}) \mathrm{T} \).The cross product for vectors \( \mathbf{a} = a_1 \hat{\mathbf{i}} + a_2 \hat{\mathbf{j}} + a_3 \hat{\mathbf{k}} \) and \( \mathbf{b} = b_1 \hat{\mathbf{i}} + b_2 \hat{\mathbf{j}} + b_3 \hat{\mathbf{k}} \) is:\[ \mathbf{a} \times \mathbf{b} = (a_2b_3 - a_3b_2) \hat{\mathbf{i}} - (a_1b_3 - a_3b_1) \hat{\mathbf{j}} + (a_1b_2 - a_2b_1) \hat{\mathbf{k}} \]Applying to our vectors:\[ \mathbf{v} \times \mathbf{B} = \begin{vmatrix} \hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \ 3 \times 10^5 & 2 \times 10^5 & 0 \ 2 & 0 & 3 \end{vmatrix} = (2 \times 10^5 \cdot 3 - 0 \cdot 0 ) \hat{\mathbf{i}} - (3 \times 10^5 \cdot 3 - 0 \cdot 2) \hat{\mathbf{j}} + (3 \times 10^5 \cdot 0 - 2 \times 10^5 \cdot 2) \hat{\mathbf{k}} \]\[ = (6 \times 10^5) \hat{\mathbf{i}} - (9 \times 10^5) \hat{\mathbf{j}} - (4 \times 10^5) \hat{\mathbf{k}} \].

Step 3: Compute the Force \( \mathbf{F} \)

Since \( \mathbf{F} = q (\mathbf{v} \times \mathbf{B}) \), the force \( \mathbf{F} \) can be rewritten using the specific charge \( \frac{q}{m} \) as:\[ \mathbf{F} = m \cdot \left(\frac{q}{m}\right) \cdot (\mathbf{v} \times \mathbf{B}) \].Given that we are working with specific charge \( 9.6 \times 10^7 \mathrm{C} \mathrm{kg}^{-1} \), we rearrange and find the acceleration vector \( \mathbf{a} \):\[ \mathbf{a} = \frac{\mathbf{F}}{m} = \left(\frac{q}{m}\right) (\mathbf{v} \times \mathbf{B}) = 9.6 \times 10^{7} \times (6 \hat{\mathbf{i}} - 9 \hat{\mathbf{j}} - 4 \hat{\mathbf{k}}) \times 10^5 \].

Step 4: Simplify and Compare with Options

Simplify the expression for acceleration:\[ \mathbf{a} = (6 \hat{\mathbf{i}} - 9 \hat{\mathbf{j}} - 4 \hat{\mathbf{k}}) \times 9.6 \times 10^{12} \].Compare this expression to the given options.The correct match for this is option (c): \( (6 \hat{\mathbf{i}} - 9 \hat{\mathbf{j}} - 4 \hat{\mathbf{k}}) \times 9.6 \times 10^{12} \).

Key Concepts

Cross Product

The cross product is a mathematical operation that takes two vectors and returns a third vector that is perpendicular to the plane containing the first two. In physics, the cross product is fundamental in calculating cross-magnetic forces. You will often encounter it when working with magnetic fields, like in the case of the Lorentz force.

In our problem, we used the cross product to find the result of a proton velocity vector multiplied by a magnetic field vector. The formula for a cross product, \[ \mathbf{a} \times \mathbf{b} = (a_2b_3 - a_3b_2) \hat{\mathbf{i}} - (a_1b_3 - a_3b_1) \hat{\mathbf{j}} + (a_1b_2 - a_2b_1) \hat{\mathbf{k}} \]might seem a bit theoretical, but it's actually easy to apply step by step.

• Calculate for each component
• Combine the calculated components

In our case, the specific cross product operation gives us \(6 \times 10^5 \hat{\mathbf{i}} - 9 \times 10^5 \hat{\mathbf{j}} - 4 \times 10^5 \hat{\mathbf{k}} \), giving clarity to the force experienced by the proton due to the magnetic field.

Magnetic Field

A magnetic field is a vector field that describes the magnetic influence on moving electric charges, electric currents, and magnetic materials. When a charged particle moves through a magnetic field, it experiences a force due to this field, known as the Lorentz force.

In physics problems, you'll often use magnetic fields to understand how charged particles like electrons and protons behave when subjected to these forces. It's important to remember that the magnetic field vector, denoted as \( \mathbf{B} \), consists of components along the three Cartesian directions (i.e., \( \hat{\mathbf{i}}, \hat{\mathbf{j}}, \hat{\mathbf{k}} \)).In our exercise, the magnetic field vector \((2 \hat{\mathbf{i}} + 3 \hat{\mathbf{k}}) \mathrm{T} \) interacts with the proton's velocity to create a force. The resulting interaction is captured mathematically by the cross product, which is fundamental in understanding how the motion vector and the magnetic vector interact to produce a resultant force.

• Fields exert forces on charged particles
• Perpendicularity is key in cross product operations
• This force causes circular or helical motion, vital in electromagnetism

Understanding magnetic fields in this context is critical for broader applications in electromagnetism and current technologies, such as MRI machines, which use magnetic fields effectively.

Specific Charge

The concept of specific charge refers to the charge-to-mass ratio of a particle, denoted as \( \frac{q}{m} \). This parameter is crucial in predicting how a particle will move in electric and magnetic fields.

If you've ever wondered why very small particles like protons and electrons have different accelerations despite having charge, the specific charge explains this well. It's essentially how much charge a particle carries per unit mass, and it's vital in calculating motion, especially under magnetic fields.In this textbook example, the specific charge is given as \(9.6 \times 10^7 \mathrm{C} \mathrm{kg}^{-1} \), which helps us find the proton's acceleration. The larger the specific charge, the more significantly the particle responds to electromagnetic fields, making this concept quite fitting for problems involving magnetic fields.

• Helps determine particle acceleration
• Key metric in experiments and particle physics
• A factor in practical applications like mass spectrometry

Grasping specific charge allows students to connect the dots between charge, mass, and motion, providing a deeper and intuitive understanding of physical phenomena involving charged particles.