【People's Education Press】High School Physics: Elective Course II: The Effect of Magnetic Fields on Current and Charges – Bridging Macroscopic to Microscopic Scales

From the Three Gorges Dam to Cathode Ray Tubes: A Dialogue Across Scales of Force

In this lesson, we will embark on a journey frommacroscopic mechanicstomicroscopic dynamicsof thought. Imagine the rushing water at the Three Gorges Dam driving a massive generator. The core physical process involves conductors moving in a magnetic field to generate current, while the resistance experienced by the current in the field is fundamentally the macroscopic accumulation of Lorentz forces acting on countless microscopic charges.Lorentz forceon a macroscopic scale. Conversely, in old-style televisions,scanningtechnology uses magnetic fields to control electron beams: one scan from top to bottom row is called a frame.

Core Physical Laws

Ampere Force (Ampere Force): $F = B I l$ . The force reaches its maximum when the current is perpendicular to the uniform magnetic field.
Lorentz Force (Lorentz Force): $F = q v B$ . It is the microscopic essence of the Ampere force. In a uniform magnetic field, if a charge moves in a circular path, then $q v B = m \frac{v^{2}}{r}$ .
Motion Laws: The period of a charged particle undergoing uniform circular motion in a uniform magnetic field is $T = \frac{2 π m}{q B}$ , surprisingly, independent of the orbital radius and the speed of motion.

Reflection and Forward Thinking

How do power plant generators achieve energy conservation? Preliminary experimental results indicate that the magnetic field generated by induced current always opposes the change in magnetic flux that caused it, known asLenz's Law.

The massive power units at the Three Gorges Dam convert mechanical energy into electrical energy through the interaction of magnetic fields with current; the source of this macroscopic force lies in the collective action of microscopic particles.

QUESTION 1

In the magnetic field shown in Figure 1.1-9, there is a current-carrying wire oriented perpendicular to the magnetic field direction. If the current

I

is directed to the right, and the magnetic induction intensity

B

is directed into the plane of the paper, what is the direction of the Ampere force according to the left-hand rule?

Upward

Downward

Into the page

Out of the page

QUESTION 2

Regarding the definition formula for magnetic induction intensity

B = F / I l

and the force calculation formula

F = I l B

, which of the following statements is correct:

F = 0

, then the magnetic induction intensity at this location

B

must be zero

B

is determined by

F

and

I

and

l

together

B = F / I l

is defined by a ratio method, and

B

its properties are determined by the magnetic field itself

When the current direction is parallel to the magnetic field direction, the formula

F = I l B

still applies

❌ Incorrect

Note

F = 0

may occur because the current is parallel to the magnetic field, and

B

its definition uses a ratio method, independent of the magnitude of the force.

QUESTION 3

In the experiment shown in Figure 1.1-12, the lower end of a flexible spring just touches mercury. What happens to the spring after being energized?

Remains stationary

Continuously contracts and moves away from the mercury surface

Vibrates up and down

Continuously deflects to the left

QUESTION 4

In Figure 1.1-10, if the conductor rod

a b

is placed on a smooth inclined plane with an angle of

θ

, and the magnetic field

B

is vertically upward, and the current flows from

a

b

. What is the direction of the Ampere force on the conductor rod?

Up the incline

Perpendicular to the incline, upward

Horizontally to the right

Horizontally to the left

✅ Correct!

Correct. According to the left-hand rule, magnetic field lines go upward through the palm, and the fingers point into the page (or according to the flow from

a

b

to the flow), the thumb points horizontally to the right.

QUESTION 5

As shown in Figure 1-1, there is a straight current-carrying wire in a curved magnetic field

a b

. If the current flows from

a

b

, how will the wire move?

Moves only upward

Rotates while moving downward

Remains stationary

Moves in uniform circular motion

Experimental Investigation: Measuring Magnetic Induction Intensity Using a Current Balance

Measuring Microscopic Field Strength via Macroscopic Force Equilibrium

As shown in Figure 1.1-11, the right arm of the current balance suspends a rectangular coil with n turns

n

, and a base length of

l

The lower edge of the coil is located within a uniform magnetic field. When a current

I

is passed through the coil, the mass

m

on the left pan is adjusted to balance the scale.

Please derive the expression for the magnetic induction intensity

n

and

m

and

l

and

I

in terms of

B

and calculate: when

n = 9

l = 10.0 cm

I = 0.10 A

m = 8.78 g

when (taking

g = 9.8 {m/s}^{2}

), what is the magnetic induction intensity?

Answer:
According to equilibrium conditions, the Ampere force

F = n B I l

is balanced against the gravitational difference

m g

. Therefore,

B = \frac{m g}{n I l}

. Substituting data:

B = \frac{0.00878 \times 9.8}{9 \times 0.10 \times 0.10} \approx 0.956 T

Deep Reflection: If we double the current in the coil while keeping the magnetic induction intensity constant, can the balance still be maintained? If not, how should the weights be adjusted?

Answer:
It cannot remain balanced. After doubling the current, the Ampere force

F

becomes twice the original value. At this point, the force on the right arm increases, requiring an increase in the mass of the left-side weights. The added mass should equal the original weight mass

m

(assuming the weights were used solely to counteract the Ampere force when originally balanced).