Peoples Physics Book Version-3

14.1. The Big Idea http://www.ck12.org

The direction of the induced current is determined as follows: the current will flow so as to generate a magnetic
field thatopposesthe change in flux. This is called Lenz’s Law. Note that the electromotive force described above
is not actually a force, since it is measured in Volts and acts like an induced potential difference. It was originally
called that since it caused charged particles to move — henceelectromotive— and the name stuck (it’s somewhat
analogous to calling an increase in a particle’s gravitational potential energy difference a gravitomotive force).

Since only a changing flux can produce an induced potential difference, one or more of the variables in equation [5]
must be changing if the ammeter in the picture above is to register any current. Specifically, the following can all
induce a current in the loops of wire:

• Changing the direction or magnitude of the magnetic field.


• Changing the loops’ orientation or area.


• Moving the loops out of the region with the magnetic field.

Example 1: Find the Magnetic Field

Question: An electron is moving to the east at a speed of 1. 8 × 106 m/s. It feels a force in the upward direction
with a magnitude of 2.2? 10−^12 N. What is the magnitude and direction of the magnetic field this electron just passed
through?

Answer: There are two parts to this question, the magnitude of the electric field and the direction. We will first focus
on the magnitude.

To find the magnitude we will use the equation

FB=qvBsinθ

We were given the force of the magnetic field( 2 .2? 10−^12 N)and the velocity that the electron is traveling( 1. 8 ×
106 m/s). We also know the charge of the electron( 1. 6 × 10 −^19 C). Also, because the electron’s velocity is
perpendicular to the field, we do not have to deal with sinθbecause sinθof 90 degrees is 1. Therefore all we
have to do is solve for B and plug in the known values to get the answer.

FB=qvBsinθ

Solving for B:

B=

FB

qvsinθ

Now, plugging the known values we have

B=

FB

qvsinθ

=

2 .2? 10−^12 N

1. 6 × 10 −^19 C× 1. 8 × 106 m/s× 1

= 7 .6T

Now we will find the direction of the field. We know the direction of the velocity (east) and the direction of the force
due to the magnetic field (up, out of the page). Therefore we can use the second right hand rule (we will use the left
hand, since an electron’s charge is negative). Point the pointer finger to the right to represent the velocity and the
thumb up to represent the force. This forces the middle finger, which represents the direction of the magnetic field,
to point south. Alternatively, we could recognize that this situation is illustrated for apositiveparticle in the right
half of the drawing above; for a negative particle to experience the same force, the field has to point in the opposite
direction: south.