Problem 14 - Master Resource Book in JEE Main Physics

Question

An electron is shot in steady electric and magnetic fields such that its velocity \(v\), electric field \(E\) and magnetic field \(B\) are mutually perpendicular. The magnitude of \(E\) is \(1 \mathrm{Vcm}^{-1}\) and that of \(B\) is \(2 \mathrm{~T}\). Now if it so happens that the Lorentz (magnetic) force cancels the electrostatic force on the electron, then the velocity of the electron is (a) \(50 \mathrm{~ms}^{-1}\) (b) \(2 \mathrm{cms}^{-1}\) (c) \(0.5 \mathrm{cms}^{-1}\) (d) \(200 \mathrm{cms}^{-1}\)

Step-by-Step Solution

Answer

The velocity of the electron is \( 50 \text{ m/s} \), matching option (a).

Step 1: Understanding Lorentz Force and Electrostatic Force

The problem involves finding the velocity of an electron where the Lorentz force due to the magnetic field and the electrostatic force due to the electric field cancel each other. The Lorentz force for a charge is given by \( F_B = q(v \times B) \), where \( q \) is the charge, \( v \) is the velocity, and \( B \) is the magnetic field. The electrostatic force is given by \( F_E = qE \), where \( E \) is the electric field. Since these forces cancel, they have equal magnitudes: \( | F_E | = | F_B | \).

Step 2: Setting up the Equation for Forces

Write the equation for the scenario where the magnitudes of the forces are equal: \[ qE = qvB \] Here, \( q \) is the electron charge, \( v \) is the velocity we need to find, \( E \) is given as \( 1 \text{ V/cm} \) which is \( 100 \text{ V/m} \), and \( B \) is given as \( 2 \text{ T} \). Since the charge \( q \) is present on both sides, it cancels out, simplifying to: \[ E = vB \]

Step 3: Solving for Velocity

Re-arrange the equation \( E = vB \) to solve for velocity \( v \): \[ v = \frac{E}{B} \] Substitute the given values into the equation: \[ v = \frac{100 \text{ V/m}}{2 \text{ T}} = 50 \text{ m/s} \].

Step 4: Selecting the Correct Answer

Compare the calculated velocity \( v = 50 \text{ m/s} \) to the provided options. Converting \( 50 \text{ m/s} \) to \( \text{ cm/s} \) (since the options are given in \( \text{ cm/s} \), except option (a)): \[ 50 \text{ m/s} = 5000 \text{ cm/s} \]. Thus, option (a) corresponds to the correct velocity magnitude.

Key Concepts

Electric and Magnetic Fields

Electric fields and magnetic fields are crucial concepts in electromagnetism. An electric field exists around electric charges, influencing other charges nearby, much like how the Earth’s gravitational field affects objects.
An electric field exerts force on charged particles, pushing them along or against the field, depending on their charge sign. It's measured in volts per meter (V/m). In our scenario, the electric field is specified as 1 V/cm, or 100 V/m in standard units.
Magnetic fields, on the other hand, are generated by moving charges or magnets. They affect moving charges, exerting a force perpendicular to both the field and the charge's velocity. This field is measured in teslas (T), and in our problem, it is stated as 2 T.
Understanding how these fields interact with charged particles is key to solving problems like the one involving Lorentz force.

Electron Motion

Electrons, being negatively charged particles, are influenced by electric and magnetic fields. When placed in an electric field, electrons experience a force that tends to move them against the field lines due to their negative charge.
In a magnetic field, the force depends on the velocity of the electron and is perpendicular to both the velocity and the magnetic field lines. This type of interaction is described by the Lorentz force, which can cause the electron to spiral, rotate, or follow some other complex path, depending on initial conditions.

• A steady electron in a magnetic field experiences no force if it’s at rest or moving parallel to the field.
• If moving perpendicular to the field, it undergoes maximum force.

The specifics of these motion tendencies are pivotal when analyzing situations, especially under the given conditions of perpendicular fields.

Force Cancellation

When we talk about force cancellation, we mean a situation where two opposing forces on an object balance each other out, resulting in no net force acting on the object.
In our electron scenario, the Lorentz force, coming from the interaction with the magnetic field, and the electrostatic force due to the electric field, cancel each other. The Lorentz force's magnitude equals the magnitude of the electrostatic force but acts in the opposite direction.
This occurs under the condition: \[ qE = qvB \] Here, each force's strength depends both on the field and the velocity of the electron. Since the electric and magnetic fields are mutually perpendicular, and they affect the moving electron differently, equating their magnitudes allows us to find specific electron behavior under these fields.
This phenomenon not only determines the specific velocity of the electron needed for equilibrium but also highlights the elegance of balancing forces within electromagnetic fields.

Velocity Calculation

To deduce the specific velocity at which an electron experiences equal and opposing electric and magnetic forces, we use the condition for force cancellation, \[ qE = qvB \]By simplifying and solving for velocity, we find that the charge cancels from both sides, giving us: \[ v = \frac{E}{B} \]With provided field values, E = 100 V/m and B = 2 T, substituting gives: \[ v = \frac{100 \text{ V/m}}{2 \text{ T}} = 50 \text{ m/s} \]
This velocity indicates the electron's speed when the forces precisely counterbalance each other, meaning it neither accelerates nor decelerates.

• Rearranging equations allows insight into interactions between particles and fields.
• This specific result is pivotal when designing systems needing precise control over charge movement, such as in magnetic traps or accelerators.

The right understanding of these calculations is necessary to grasp the concepts at play, often encountered in advanced applications of electromagnetism.