Problem 3 - Master Resource Book in JEE Main Physics

Question

In a chamber, a uniform magnetic field of \(6.5 \mathrm{G}\) \(\left(1 \mathrm{G}=10^{-4} \mathrm{~T}\right)\) is maintained. An electron is shot into the field with a speed of \(4.8 \times 10^{6} \mathrm{~m} / \mathrm{s}\) normal to the field explain why the path of the electron is a circle. If \(\left(e=1.6 \times 10^{-19} \mathrm{C}, m_{e}=9.1 \times 10^{-31} \mathrm{~kg}\right)\), then obtain the frequency of revolution of the electron in its circular orbit. (a) \(6 \times 10^{6} \mathrm{~Hz}\) (b) \(18.18 \times 10^{6} \mathrm{~Hz}\) (c) \(10.10 \times 10^{6} \mathrm{~Hz}\) (d) \(12.10 \times 10^{6} \mathrm{~Hz}\)

Step-by-Step Solution

Answer

The correct answer is (b) \(18.18 \times 10^6 \mathrm{~Hz}\).

Step 1: Determine the Magnetic Force on the Electron

The magnetic force acting on the electron moving perpendicular to the magnetic field is given by the equation \( F = qvB \), where \( q \) is the charge of the electron, \( v \) is the velocity of the electron, and \( B \) is the magnetic field strength. Substituting the values, \( q = 1.6 \times 10^{-19} \mathrm{~C} \), \( v = 4.8 \times 10^6 \mathrm{~m/s} \), and \( B = 6.5 \times 10^{-4} \mathrm{~T} \), we get:\[ F = (1.6 \times 10^{-19}) \cdot (4.8 \times 10^6) \cdot (6.5 \times 10^{-4}) \]Calculating this gives the magnetic force \( F \).

Step 2: Understand Why the Path is Circular

The magnetic force acts as a centripetal force, causing the electron to move in a circular path. This is because the force is always perpendicular to the velocity of the electron, which changes the direction of the velocity without changing its magnitude, leading to a circular motion.

Step 3: Calculate the Radius of the Circular Path

The radius \( r \) of the circular path can be found using the formula for centripetal force: \( F = \frac{mv^2}{r} \). Setting the magnetic force equal to this expression gives \( qvB = \frac{mv^2}{r} \). Solving for \( r \), we find:\[ r = \frac{mv}{qB} \]However, to find the frequency of revolution, we'll use the values calculated so far.

Step 4: Derive the Frequency of Revolution

The frequency of revolution, \( f \), is found using the formula \( f = \frac{qB}{2\pi m} \), which comes from equating the centripetal force to the magnetic force and using \( T = \frac{2\pi r}{v} \) for period. Substitute values: \( q = 1.6 \times 10^{-19} \mathrm{~C} \), \( B = 6.5 \times 10^{-4} \mathrm{~T} \), and \( m = 9.1 \times 10^{-31} \mathrm{~kg} \):\[ f = \frac{1.6 \times 10^{-19} \times 6.5 \times 10^{-4}}{2\pi \times 9.1 \times 10^{-31}} \]Compute this expression to find \( f \).

Step 5: Compute the Frequency Value

Calculate the expression for frequency:\[ f = \frac{1.6 \times 6.5}{2\pi \times 9.1} \times 10^{6} \]Approximating the numbers gives:\[ f \approx 18.18 \times 10^6 \text{ Hz} \]Therefore, the correct answer is option (b).

Key Concepts

Magnetic Force

When an electron moves through a magnetic field, it experiences a magnetic force. This force is given by the equation \( F = qvB \), where \( q \) is the charge of the electron, \( v \) is its velocity, and \( B \) is the strength of the magnetic field. The magnetic force acts perpendicular both to the direction of the velocity of the electron and to the magnetic field. This unique property causes the electron to follow a circular path. Key points to remember about magnetic force:

• It is dependent on the charge of the particle.
• It depends on the velocity at which the particle moves.
• It’s always perpendicular to both velocity and magnetic field direction.

Understanding how magnetic force operates is crucial for analyzing the motion of charged particles like electrons in magnetic fields.

Circular Motion

In a magnetic field, the electron travels in a circular path due to the magnetic force acting as the centripetal force. Centripetal force is the force that keeps an object moving in a circular path and is always directed towards the center of the circle. When the magnetic force becomes the centripetal force equation \( F = \frac{mv^2}{r} \), it ensures the electron's velocity continuously changes direction, maintaining its circular trajectory. Despite the tortuous path, the speed remains constant as no work is done by the magnetic force. Points to consider:

• Only the direction of velocity changes, not the magnitude.
• The radius of the circle is influenced by the electron's mass, speed, charge, and magnetic field strength.
• Circular motion showcases how magnetic forces disrupt linear paths in charged particles.

This circular motion is a classic demonstration of the interplay between velocity, force, and radius in magnetic fields.

Frequency of Electron Revolution

The frequency of revolution of an electron in a magnetic field is derived from the relationship between the magnetic force and centripetal acceleration. The equation \( f = \frac{qB}{2\pi m} \) stems from analyzing the period of one circular orbit. Here, \( f \) represents frequency, \( q \) is the charge, \( B \) is the magnetic field strength, and \( m \) is the mass of the electron. Understanding frequency calculations is valuable because:

• The frequency signifies how often an electron completes a cycle around its circular path in one second.
• It is influenced by the intrinsic properties of the electron (mass and charge) and external factors like field strength.
• This frequency remains constant as it does not depend on the actual speed of the electron in the field.

Through controlled calculations and understanding of these values, predicting the electron’s behavior in magnetic fields becomes straightforward.