## Wednesday, July 1, 2020

### UNIZOR.COM - Physics4Teens - Electromagnetism

Notes to a video lecture on http://www.unizor.com

Problems on
Electromagnetic Induction

Problem A

The following experiment is conducted in the space with Cartesian coordinates.

Two infinitely long parallel wires in XY-plane are parallel to X-axis, one at Y-coordinate y=a and another at Y-coordinate y=−a (assuming a is positive).

These two wires are connected by a third wire positioned along the Y-coordinate between points (0,a) and (0,−a).

The fourth wire, parallel to the third one, also connects the first two,
but slides along the X-axis, always maintaining its parallel position
to Y-axis. The X-coordinate of its position is monotonously increasing
with time t according to some rule x=x(t).

All four wires are made of the same material and the same cross-section with the electrical resistance of a unit length r.

There is a uniform magnetic field of intensity B with field lines parallel to Z-axis.

Initial position of the fourth wire coincides with the third one, that is x(0)=0.

What should the function x(t) be to assure the generation of the same constant electric current I0 in the wire loop?

What is the speed of the fourth rod at initial time t=0?

Solution

The area of a wire frame is changing with time:

S(t)=2a·x(t).

Magnetic flux through this wire frame is

Φ(t)=B·S(t).

Therefore, the magnitude of the generated electromotive force or voltage U(t), as a function of time, is

U(t) = dΦ(t)/dt = 2B·a·x'(t)

The resistance R(t) of the wire loop, as a function of time, is a product of a resistance of a unit length of a wire r by the total length of all four sides of a wire rectangle L=4a+2x(t).

R(t) = r·L = 2r·[2a+x(t)]

The electric current I(t) in a wire loop, according to the Ohm's Law, is

I(t) = U(t)/R(t) =

= 2B·a·x'(t) /
{2r·[2a+x(t)]} =

= B·a·x'(t) /
{[2a+x(t)]}

This current has to be constant and equal to I0. This leads us to a differential equation I0=I(t)

I0 = B·a·x'(t) / {[2a+x(t)]}

Simplifying this equation, obtain

I0·r / (B·a) = x'(t) / [2a+x(t)]

[I0·r / (B·a)]·dt =

=
d
[2a+x(t)] / [2a+x(t)]

Integrating,

[I0·r / (B·a)]·t + C = ln(2a+x(t))

2a + x(t) = C·eI0·r·t / (B·a)

Since x(0)=0, C=2a

x(t) = 2a·[eI0·r·t / (B·a) − 1]

Speed of the motion of the fourth wire along the X-axis is

x'(t) = 2a·eI0·r·t/(B·a)·[I0·r/(B·a)]

or

x'(t) = eI0·r·t/(B·a)·[2I0·r/B]

At time t=0 the initial speed is

x'(0) = 2I0·r / B

Problem B

A rectangular wire frame in a space with Cartesian coordinates rotates with variable angular speed ω(t) in a uniform magnetic field B.

In the beginning at t=0 the wire frame is at rest, ω(0)=0.
As the time passes, the angular speed is monotonously increasing from
initial value of zero to some maximum. This models turning the rotation
on.

The axis of rotation is Z-axis.

The initial position of a wire frame is that its plane coincides with XZ-plane.

The magnetic field lines are parallel to X-axis.

So the angle between the magnetic field lines and the wire frame plane φ(t) at t=0 equals to zero.

The sides of a wire frame parallel to Z-axis (those, that cross the magnetic field lines) have length a, the other two sides have length b.

Determine the generated electromotive force (EMF) in this wire frame as a function of time t.

Solution

The angle φ(t) between the magnetic field lines and the
wire frame plane is changing with time. In the initial position, when
the wire frame coincides with XZ-plane, this angle is 0°, since magnetic
field lines are stretched along the X-axis.

As the wire frame rotates with variable angular speed ω(t), the angle between magnetic field lines and the wire frame plane φ(t) and the angular speed ω(t) are related as follows

dφ/dt = φ'(t) = ω(t)

This is sufficient to determine the value of φ(t) by integration of the angular speed on a time interval [0,t]

φ(t) = [0,t] ω(τ)·dτ

Magnetic field flux Φ(t) flowing through a wire frame depends on the intensity of the magnetic field B, area of a wire frame S=a·b and angle φ(t) the plane of a wire frame makes with magnetic field lines.

Φ(t) = B·S·sin(φ(t))

The electromotive force (voltage) U(t) generated by rotating wire frame equals to a rate of change (first derivative) of the magnetic field flux

U(t) = dΦ(t)/dt =

= B·S·cos(φ(t))·φ'(t) =

= B·S·cos(φ(t))·ω(t)

where angle φ(t) can be obtained by an integration of an angular speed presented above.