   Chapter 19, Problem 14P

Chapter
Section
Textbook Problem

Figure P19.14a is a diagram of a device called a velocity selector, in which particles of a specific velocity pass through undeflected while those with greater or lesser velocities are deflected either upwards or downwards. An electric field is directed perpendicular to a magnetic field, producing an electric force and a magnetic force on the charged particle that can be equal in magnitude and opposite in direction (Fig. P19.14b) and hence cancel. Show that particles with a speed of v = E/B will pass through the velocity selector undeflected. Figure P19.14

To determine

The particles with a speed of v=EB will pass through the velocity selector un-deflected.

Explanation

Given info: The velocity of the positively charged particle is directed on the plane of the page towards right. The magnetic field is directed into the page. Electric field is directed on the plane of the page downwards.

The magnetic force on a charged particle is given by,

Fm=qvBsinθ      (1)

• q is the charge on the particle
• v is the velocity of the particle
• B is the magnetic field
• θ is the angle between the velocity and the magnetic field

Since the velocity of the proton is perpendicular to the magnetic field, the magnetic force will be maximum having a value,

Fm=qvB

Right hand rule: Consider a positive charge having a velocity v, moving in a magnetic field B. The direction of the magnetic force on the positive charge can be found out by using right hand rule. When the fingers of the right hand are curled from the direction of v towards the direction of B, the magnetic force will direct in the direction of thumb

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