Book edition 10th Edition

Author(s) David Halliday

Pages 1328 pages

ISBN 9781118230718

Answers without the blur.

Just sign up for free and you're in.

Short Answer

In Fig. 28-58, an electron of mass m, charge -e, and low (negligible) speed enters the region between two plates of potential difference V and plate separation d, initially headed directly toward the top plate. A uniform magnetic field of magnitude B is normal to the plane of the figure. Find the minimum value of B, such that the electron will not strike the top plate.

The minimum value of B such that the electron will not strike the top plate is
$B = \sqrt{\frac{m V}{2 e d^{2}}}$

See the step by step solution

Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1: Given

The mass of electron is m.
The charge on electron is -e.
Potential difference across the plates is V.
The distance of plate separation is d.
The magnitude of magnetic field is B.

Step 2: Determine the concept and the formulas

Consider the equations of electric force and magnetic force. The magnetic force opposes the motion of electron, so to prevent the electron from striking, the magnetic force must be greater than the electric force.

Formulae:

$F_{B} = q v B$
$F_{E} = \frac{q V}{d}$
$K E = \frac{1}{2} m v^{2}$
$P E = q V$

Step 3: Calculate the minimum value of B such that the electron will not strike the top plate.

Consider the equation for potential energy of electron just before it strikes the top plate as,

$U = q E = \frac{e V}{d}$

Now, according to the conservation of energy principle is as follows:

KE=PE

So,

$\frac{1}{2} m v^{2} = e V$

Rearranging for velocity derive the equation as:

$v = \sqrt{\frac{2 e V}{m}}$

Now, consider the equation for magnetic force as

$F_{B} = q v B = e v B$

Substituting for the velocity, we get

$F_{B} = e B \sqrt{\frac{2 e V}{m}}$

Now, to prevent the electron from striking the top plate, F_B must be greater than F_E.

So, to find the minimum magnetic field, let us assume those to be equal.

Hence,

F_B=F_E

$e B \sqrt{\frac{2 e V}{m}} = \frac{e V}{d}$

Rearranging it for magnetic field, solve as:

$\begin{array}{l} B = \frac{V}{d} \times \sqrt{\frac{m}{2 e V}} \\ B = \sqrt{\frac{m V}{2 e d^{2}}} \end{array}$

Find Study Materials for

Create Study Materials

Select your language

Fundamentals Of Physics

Answers without the blur.

Short Answer

Step by Step Solution

Step 1: Given

Step 2: Determine the concept and the formulas

Step 3: Calculate the minimum value of B such that the electron will not strike the top plate.

Most popular questions for Physics Textbooks

Want to see more solutions like these?

Recommended explanations on Physics Textbooks

Select your language

Fundamentals Of Physics

Answers without the blur.

Short Answer

Step by Step Solution

Step 1: Given

Step 2: Determine the concept and the formulas

Step 3: Calculate the minimum value of B such that the electron will not strike the top plate.

Most popular questions for Physics Textbooks

Want to see more solutions like these?

Recommended explanations on Physics Textbooks

Work Energy and Power

Radiation

Astrophysics

Physics of Motion

Scientific Method Physics

Fluids

Torque and Rotational Motion

Nuclear Physics

Fields in Physics

Measurements

Space Physics

Electricity

Physical Quantities and Units

Medical Physics

Electrostatics

Further Mechanics and Thermal Physics

Mechanics and Materials

Engineering Physics

Energy Physics

Force

Particle Model of Matter

Waves Physics

Magnetism

Modern Physics

Atoms and Radioactivity

Dynamics

Electricity and Magnetism

Conservation of Energy and Momentum

Fundamentals of Physics

Kinematics Physics

Circular Motion and Gravitation

Linear Momentum

Rotational Dynamics

Oscillations

Translational Dynamics

Geometrical and Physical Optics

Thermodynamics

Magnetism and Electromagnetic Induction

Electromagnetism

Electric Charge Field and Potential

Turning Points in Physics