Problem 9 - Master Resource Book in JEE Main Physics

Question

Cathode rays of velocity \(10^{6} \mathrm{~ms}^{-1}\) describe an approximate circular path of radius \(1 \mathrm{~m}\) in an electric field of \(500 \mathrm{~V} \mathrm{~cm}^{-1}\). If the velocity of cathode rays is doubled, the value of electric field needed so that the rays describe the same circular path is (a) \(1000 \mathrm{~V} \mathrm{~cm}^{-1}\) (b) \(1500 \mathrm{~V} \mathrm{~cm}^{-1}\) (c) \(2000 \mathrm{~V} \mathrm{~cm}^{-1}\) (d) \(500 \mathrm{~V} \mathrm{~cm}^{-1}\)

Step-by-Step Solution

Answer

(c) 2000 \(\text{V cm}^{-1}\)

Step 1: Understand the Problem

To find the new electric field required to maintain the same circular path when the velocity of cathode rays is doubled, start by considering the relationship between the original and new electric field and velocity.

Step 2: Initial Relation Between Electric Field and Velocity

Initially, the velocity of the cathode rays is \(10^6 \mathrm{~ms}^{-1}\) and the electric field is \(500 \mathrm{~V} ~ \mathrm{cm}^{-1}\). The force acting on the cathode rays is centrifugal and is countered by the electric force, giving the relationship that the electric field \( E \) is proportional to the square of the velocity, based on the centripetal force requirement.

Step 3: Calculate New Velocity

The velocity of the cathode rays is doubled, so the new velocity \( v' = 2 \times 10^6 \mathrm{~ms}^{-1}\).

Step 4: Establish Proportionality for Electric Field

For a particle moving in a circle under the influence of an electric field, the electric field \( E \) required is proportional to the square of the velocity: \( E \propto v^2 \). Thus, \( E' = E \left(\frac{v'}{v}\right)^2 \).

Step 5: Substitute Values and Solve

Substitute the known values into the proportionality equation: \[ E' = 500 \mathrm{~V} ~ \mathrm{cm}^{-1} \left(\frac{2 \times 10^6}{1 \times 10^6}\right)^2 = 500 \mathrm{~V} ~ \mathrm{cm}^{-1} \times 4 = 2000 \mathrm{~V} ~ \mathrm{cm}^{-1} \]

Step 6: Choose the Correct Answer

From the calculated value, the new electric field required is \(2000 \mathrm{~V} ~ \mathrm{cm}^{-1}\). Therefore, the correct choice is (c) \(2000 \mathrm{~V} ~ \mathrm{cm}^{-1}\).

Key Concepts

Centripetal Force

When a charged particle like a cathode ray moves in a circular path, it experiences a special kind of force known as centripetal force. This force is essential because it keeps the particle following a curved path, instead of moving off in a straight line.
In the case of cathode rays moving in an electric field, this centripetal force is a result of the electric force acting on the particle. The amount of force needed to keep the particle in a circular motion depends on two main factors:

• The mass of the particle
• Square of its velocity

The centripetal force is given by the formula:\[ F_c = \frac{mv^2}{r} \]where \( m \) is the mass of the particle, \( v \) is its velocity, and \( r \) is the radius of the circular path. Understanding this relationship is key to figuring out why changes in velocity can significantly alter the amount of electric field required to maintain the same circular path.

Cathode Rays

Cathode rays are streams of electrons observed in vacuum tubes. These electrons move at high speeds and exhibit interesting properties when subjected to electric and magnetic fields. They can be deflected by these fields, resulting in curved paths.
A cathode ray's path can become circular when it moves through an electric field of appropriate strength, providing insight into their interaction with electric forces. Because cathode rays are negatively charged, they will move under the influence of positive and negative charges in an electric field, which can then affect their trajectory.
The motion of cathode rays in electric fields forms the basis for many practical applications, such as the operation of old television tubes and cathode-ray oscilloscopes. Grasping the behavior of cathode rays under electric influences is vital for understanding fundamental electronic principles.

Velocity and Electric Field Relationship

The relationship between a particle's velocity and the electric field it experiences is a fundamental concept when studying circular motion. In essence, the force exerted by the electric field must be equal and opposite to the centripetal force required to keep the particle moving in a circle.
As the velocity of a charged particle like a cathode ray increases, the required electric field strength also increases to provide sufficient centripetal force. Mathematically, this relationship can be expressed by assuming the electric field force equates to the centripetal force:

• Electric force: \( F_e = E imes q \)
• Centripetal force: \( F_c = \frac{mv^2}{r} \)

Equating these forces provides:\[ E = \frac{mv^2}{qr} \]This shows that as the velocity \( v \) doubles, the electric field \( E \) must increase by a factor of four to maintain the same path. Therefore, for a cathode ray moving in an electric field, understanding how changes in velocity affect the required electric field strength is crucial for accurately controlling their trajectory.