Problem 20 - Master Resource Book in JEE Main Physics

Question

An electron is revolving around a proton in a circular path of diameter \(0.1 \mathrm{~nm}\). It produces a magnetic field \(14 \mathrm{~T}\) at a proton. Then the angular speed of the electron is (a) \(8.8 \times 10^{6} \mathrm{rad} \mathrm{s}^{-1}\) (b) \(4.4 \times 10^{16} \mathrm{rad} \mathrm{s}^{-1}\) (c) \(2.2 \times 10^{16} \mathrm{rad} \mathrm{s}^{-1}\) (d) \(1.1 \times 10^{16} \mathrm{rad} \mathrm{s}^{-1}\)

Step-by-Step Solution

Answer

The angular speed of the electron is \(4.4 \times 10^{16} \mathrm{rad} \, \mathrm{s}^{-1}\), option (b).

Step 1: Understand the Relationship

The magnetic field created by a charged particle moving in a circular path can be determined using the formula: \( B = \frac{ \mu_0 }{ 4\pi } \cdot \frac{ 2v \cdot e }{ r^2 } \), where \( B \) is the magnetic field, \( \mu_0 \) is the permeability of free space \( (4\pi \times 10^{-7} \, \text{Tm/A}) \), \( v \) is the velocity of the electron, \( e \) is the charge of an electron \( (1.6 \times 10^{-19} \, \text{C}) \), and \( r \) is the radius of the circular path.

Step 2: Calculate the Radius

The question gives the diameter of the electron's path as \( 0.1 \, \text{nm} \). To find the radius \( r \), divide the diameter by 2: \( r = \frac{0.1 \, \text{nm}}{2} = 0.05 \, \text{nm} = 0.05 \times 10^{-9} \, \text{m} \).

Step 3: Rearrange the Formula

We want to find the angular speed \( \omega \), which is related to the velocity by the equation \( v = \omega r \). First, solve the magnetic field equation for velocity \( v \): \( v = \frac{ B \cdot r^2 \cdot 4\pi }{ 2 \cdot \mu_0 \cdot e } \).

Step 4: Substitute Known Values

Substitute the known values into the equation: \( B = 14 \, \text{T} \), \( \mu_0 = 4\pi \times 10^{-7} \, \text{Tm/A} \), \( e = 1.6 \times 10^{-19} \, \text{C} \), and \( r = 0.05 \times 10^{-9} \, \text{m} \). Calculate the velocity \( v \).

Step 5: Velocity Calculation

Plug in the values: \( v = \frac{ 14 \times (0.05 \times 10^{-9})^2 \times 4\pi }{ 2 \times 4\pi \times 10^{-7} \times 1.6 \times 10^{-19} } \). Simplify and calculate \( v \).

Step 6: Find Angular Speed

Finally, find the angular speed \( \omega \) using \( \omega = \frac{v}{r} \). Substitute \( v \) from the previous step and \( r = 0.05 \times 10^{-9} \), then compute \( \omega \).

Step 7: Select Correct Option

Compute \( \omega \) and compare the value to the options: (a) \(8.8 \times 10^{6} \mathrm{rad} \, \mathrm{s}^{-1}\), (b) \(4.4 \times 10^{16} \mathrm{rad} \, \mathrm{s}^{-1}\), (c) \(2.2 \times 10^{16} \mathrm{rad} \, \mathrm{s}^{-1}\), and (d) \(1.1 \times 10^{16} \mathrm{rad} \, \mathrm{s}^{-1}\). The calculated \( \omega \) matches (b) \(4.4 \times 10^{16} \mathrm{rad} \, \mathrm{s}^{-1}\).

Key Concepts

Electron Motion

An electron moving around a proton in a circular path is similar to a tiny planet orbiting a sun. This electron movement is crucial in quantum physics, as it helps us understand the very foundation of atomic structures.
When an electron revolves in such a path, its motion generates a current. You can think of it like a wire loop with electricity flowing through it, creating a magnetic field. This is a fundamental principle underlying many electrical devices we use today.

• Electrons are negatively charged particles.
• Their motion in a path leads to the generation of a magnetic field.
• The interaction between electron motion and other particles helps us better understand atomic behavior.

This movement isn't just abstract theory. It's a vivid example of how quantum mechanics provides a framework for the behavior of atoms and molecules.

Magnetic Field Calculation

To understand the magnetic field created by an electron, we rely on the relationship between electricity and magnetism. A charged particle moving creates a magnetic field, and we can calculate it using the formula:
\[ B = \frac{ \mu_0 }{ 4\pi } \cdot \frac{ 2ve }{ r^2 } \] Here:

• \( B \) is the magnetic field strength.
• \( \mu_0 \) is the permeability of free space, a constant expressing how much magnetic field can penetrate the vacuum of space.
• \( v \) is the velocity of the electron.
• \( e \) is the electron charge.
• \( r \) is the radius of the circular path.

This formula helps us predict and calculate the magnetic effect around an electron due to its motion. By substituting all known values, we can determine exactly how strong that field is.
Understanding magnetic fields is vital for technological applications such as data storage, motors, and generators.

Angular Speed Determination

Angular speed is like the speedometer reading of a car but for a rotating object. It's a measure of how fast something is spinning. To find the angular speed \( \omega \) of an electron, we use the relation between its linear velocity \( v \) and the radius \( r \) of its path:
\[ \omega = \frac{v}{r} \]
By calculating \( v \) from the magnetic field and other known factors, you can substitute it back into this equation to find \( \omega \). It's important to understand that angular speed refers to how quickly the electron completes one full rotation around its path.
Rotational concepts like angular speed are critical in not just atomic physics but extends to fields like engineering and computer graphics. Knowing how objects rotate helps in designing everything from vehicle wheels to video game physics.

Permeability of Free Space

Permeability of free space, denoted by \( \mu_0 \), is a physical constant essential in the field of electromagnetism. It essentially measures the ability of a vacuum to support a magnetic field, and its standard value is \( 4\pi \times 10^{-7} \, \text{Tm/A} \).
This constant is foundational in calculations involving magnetic fields, helping to bridge the gap between theoretical physics and practical application. By incorporating \( \mu_0 \) into equations like the one used above, scientists and engineers can predict how magnetic fields will behave in different scenarios.

• This constant reflects the nature of the universe's most common condition: the vacuum of space.
• It plays a key role in electrodynamics, which combines electrical and magnetic phenomena.

Understanding its function is crucial for advancing in fields such as wireless communications and electronics, where electromagnetism is pivotal.