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Electric Circuits

There are three important physical quantities to keep track of in an electric circuit

Electric Potential

Electric Current

Resistance (Impedance)

First, you need a conductor

as a conduit for electricity

The "conduction" electrons bounce around.

Q: What is the average speed of an electron within the metal?

A: Assuming a simple model of a 3D electron gas @ room temp:

\frac{1}{2}m_e v^2=\frac{3}{2}k_B T

v=\sqrt{\frac{3k_B T}{m_e}}\approx 10^5 \: m/s

\text{Kinetic Energy} = \text{Thermal Energy}

Electric field inside a conductor?!?

Suppose we setup an electric field inside the conductor...

Yes,charges are moving -- this is not electrostatic equilibrium!

How do you setup an electric field inside the conductor?

e.g. Discharging a capacitor

modeled here by tilting the board

So we setup an electric field inside the conductor

Q: Why are the trajectories parabolic?

A: Projectile motion

\implies \vec{a} = \frac{q_e}{m_e} \vec{E}

\implies q_e \vec{E} = m_e \vec{a}

\Delta\vec{r} = \vec{v}_0 t + \tfrac{1}{2}\vec{a}t^2

Constant acceleration in direction opposite of field,

and constant velocity perpendicular to field.

\vec{F} = m \vec{a}

So we setup an electric field inside the conductor

Q: What is the average "drift" velocity?

A: very slow ~

~ 10^{-4} m/s

{v}_{d} = a\ t

v_d = \frac{q_e \tau}{m_e} E

If the average time between collisions is

\tau

then

Then, you make a wire

\implies i_e= \frac{n_e\ A\ q_e\ \tau}{m_e} E

Conventional Current

The rate of charge flow in a conductor

I = \frac{dQ}{dt}

The SI unit for conventional current is the Ampere

A = \frac{C}{s}

Conventional Current

I = \frac{\Delta Q}{\Delta t} = \frac{q_e N_e}{\Delta t} = q_e\ i_e

Electron Current

Charging and discharging a capacitor

RC in series

(a) circuit contains a source of EMF, a resistor, an uncharged capacitor, and a switch.

Charging a Capacitor

(b) Connect the switch to (A) and start a timer (t=0)

what is the charge on the capacitor?
what is the potential difference across the capacitor?
what is the potential difference across the resistor?
What is the current through the resistor (supplied by the EMF source) ?

@ t=0 (immediately after switch is connected)

\mathcal{E}

\frac{\mathcal{E}}{R}

Charging a Capacitor

(b) Connect the switch to (A) and start a timer (t=0)

what is the charge on the capacitor?
what is the potential difference across the capacitor?
what is the potential difference across the resistor?
What is the current through the resistor (supplied by the EMF source) ?

@ t=t (some short time after switch ...)

\mathcal{E}-\frac{q}{C}

\frac{q}{C}

q(t)

\frac{\mathcal{E}}{R}-\frac{q}{RC}

Charging a Capacitor

(b) Connect the switch to (A) and start a timer (t=0)

I=\frac{\mathcal{E}}{R}-\frac{q}{RC}

\text{i.e.}\quad\frac{dq}{dt}=\frac{\mathcal{E}}{R}-\frac{q}{RC}

q(t)=C{\mathcal{E}}\left(1-e^{-\frac{t}{RC}}\right)

\text{solve for}\quad q(t)

Charging a Capacitor

(b) Connect the switch to (A) and start a timer (t=0)

q(t)=C{\mathcal{E}}\left(1-e^{-\frac{t}{RC}}\right)

what is the charge on the capacitor?
what is the potential difference across the capacitor?
what is the potential difference across the resistor?
What is the current through the resistor (supplied by the EMF source) ?

C{\mathcal{E}}

{\mathcal{E}}

@ t= (long time after switch ...)

\infty

Charging a Capacitor

(b) Connect the switch to (A) and start a timer (t=0)

q(t)=C{\mathcal{E}}\left(1-e^{-\frac{t}{RC}}\right)

I=\frac{dq}{dt}

=\frac{d}{dt}C{\mathcal{E}}\left(1-e^{-\frac{t}{RC}}\right)

=\frac{\mathcal{E}}{R}\ e^{-\frac{t}{RC}}

=I_0\ e^{-\frac{t}{\tau}}

Charging a Capacitor

(b) Connect the switch to (A) and start a timer (t=0)

q(t)=C{\mathcal{E}}\left(1-e^{-\frac{t}{RC}}\right)

I(t)=I_0\ e^{-\frac{t}{\tau}}

@just after

Capacitor draws maximum current

@long_time

Capacitor draws zero current

Charging a Capacitor


(a) The switch has been open for a long time
(b) Just after the switch is closed
(c) A long time after the switch has been closed

I_1

I_2

I_3

Check your understanding

\frac{V}{R_1}

\frac{V}{R_1+R_2}

AC-Circuits

Magnetism

Charged Particles

influence is described in terms of

The influence

and interactions

Story so far was about

Electric

Fields

3-D Electric Field Caused by two identical Point Charges

interaction is described in terms of

Electric

Forces

Charged Particles

The influence

and interactions

Today's story...

moving

3-D Electric Field Caused by two identical Point Charges

Biot-Savart Law

\vec{B}=\frac{\mu_0}{4\pi}\frac{q\ \vec{v}\ \times \ \hat{r}}{r^2}

Direction is given by

Right-Hand-Rule

Magnitude is given by

{B}=\frac{\mu_0}{4\pi}\frac{q\ v\ \text{sin}\theta }{r^2}

\mu_0=4\pi\times 10^{-7} T\ m/A

[B]=\text{T for Tesla}

Biot-Savart Law

\vec{B}=\frac{\mu_0}{4\pi}\frac{q\ \vec{v}\ \times \ \hat{r}}{r^2}

Direction is given by

Right-Hand-Rule

Magnitude is given by

{B}=\frac{\mu_0}{4\pi}\frac{q\ v\ \text{sin}\theta }{r^2}

\mu_0=4\pi\times 10^{-7} T\ m/A

[B]=\text{T for Tesla}

Biot-Savart Law

\vec{B}=\frac{\mu_0}{4\pi}\frac{q\ \vec{v}\ \times \ \hat{r}}{r^2}

•Sources are moving charges

–If velocity is zero, magnetic field is zero!

–

•Superposition principle applies

–Just like electric fields, magnetic field vectors add linearly!

–

•Biot-Savart Law

–Magnetic field strength has inverse square dependence on the distance from the moving charge

Important Highlights

Biot-Savart Law

\vec{B}=\frac{\mu_0}{4\pi}\frac{q\ \vec{v}\ \times \ \hat{r}}{r^2}

Example

Biot-Savart Law

d\vec{B}=\frac{\mu_0}{4\pi}\frac{I\ d\vec{l}\ \times \ \hat{r}}{r^2}

Magnetic Field due to a Current Segment

d\vec{B}=\frac{\mu_0}{4\pi}\frac{I\ d\vec{l}\ \times \ \hat{r}}{r^2}

\vec{B}=\frac{\mu_0}{4\pi}\frac{q\ \vec{v}\ \times \ \hat{r}}{r^2}

q \vec{v} = q \frac{d\vec{l}}{dt} = I\ d\vec{l}

Biot-Savart Law

d\vec{B}=\frac{\mu_0}{4\pi}\frac{I\ d\vec{l}\ \times \ \hat{r}}{r^2}

Magnetic Field due to a Current Segment

Magnetic Forces

\vec{F}=q\ \vec{v}\ \times\ \vec{B}

A charge q moving with velocity v relative to a magnetic field B will experience a magnetic force F given by:

Force on a charge moving relative to a magnetic field

Eg. electron in Earth's field

An electron travelling at 20% the speed of light in a direction 30 degrees North of East, passes through a region where Earth’s magnetic field is uniform, pointing North, with a magnitude of 50 µT. What is (the magnitude and direction of) the magnetic force experienced by the electron?

The electric force

is parallel to

the electric field

The magnetic force

is perpendicular to

the magnetic field

\vec{F}=q\ \vec{v}\ \times\ \vec{B}

\vec{F}=q\ \vec{E}

Electromagnetic Forces

Application: Velocity Selector

A velocity selector is a device that allows charged particles with a particular velocity to pass through, while deflecting all other charged particles.

The device operates by applying electric and magnetic forces to the particle in such a way that these

forces balance.

For the situation shown in the figure, how should an electric field be applied so that the force it applies to the particle can balance the magnetic force?

For the situation shown in the figure, what is the selected velocity?

Application: Velocity Selector

Application: Hall Effect Sensors

The electric force

is not perpendicular to

the motion

The magnetic force

is perpendicular to

the motion

W_\text{magnetic force} = 0

W_\text{electric force}\neq 0

particle speeds up or slows down

particle changes direction of motion

Work by Electromagnetic Forces

The magnetic force remains

perpendicular to the velocity and is

directed toward the center of the

circular path.

F_c = m\ \frac{v^2}{r}

q\ v\ B= m\ \frac{v^2}{r}

r= \frac{m\ v}{q\ B}

Uniform Circular Motion

m and q are intrinsic quantities

v is the speed (related to KE)

B is the magnetic flux density

Application

Charged

Particles

in

Cloud/Bubble

Chambers

If the magnetic field is directed out of the "page"--

What are the signs of the charges of the three particles?

All 3 particles have the same mass and (magnitude of) charge--

which particle is initially moving most rapidly?

All 3 particles follow a spiraling path --

Are we able to explain why?

The tracks going counter clockwise are left by negatively charged particles.

The bigger the speed, the bigger the radius (for same q, m and B)

Loss of energy (signified by loss of speed, leads to ever decreasing radius)

The mass-spectrometer

r= \frac{m\ v}{q\ B}

For fixed v, q, and B

r \propto m

Detector

particle injection

The Mass-Spectrometer

\tfrac{1}{2}mv^2=qV

r= \frac{m\ v}{q\ B}

m=\left(\frac{q\ r^2}{2\ V}\right)\ B^2

The velocity component perpendicular to the magnetic field creates circular motion,

The component of the velocity parallel to the field moves the particle along a straight line.

The resulting motion is helical.

A charged particle

moving with a velocity

neither parallel nor perpendicular

to the magnetic field.

Force on a current-carrying wire

\vec{F}=\int I\ d\vec{l}\ \times\ \vec{B}

The magnetic forces on the confined charges add up, and manifest as a force on the current carrying wire.

Force on a current-carrying wire

F= I\ L\ B \sin\theta

For a straight segment in a uniform magnetic field:

Force on a current-carrying wire

Example: Force on a cable due to Earth's field

Calculate the force per unit length on a power cable carrying 200 A directed eastwards, due to Earth's magnetic field at a location where Earth's field is 50uT pointing North. Is this force a reason for concern?

Force on a current-carrying wire

Example - force between parallel conductors:

Calculate the force per unit length between two cables running parallel to each other, carrying currents of 1.0A and 2.0 A in opposite directions.

Torque on a current-carrying loop

Consider a rectangular loop, L x W, carrying a current I, in a uniform magnetic field of strength B

F= I\ L\ B \sin\theta

The magnitude of the force on each arm of the loop is given by

These forces are in opposite directions, as given by the RHR

\implies \Sigma\vec{F} =0\quad \&\quad \Sigma\vec{\tau} \ne0

Torque on a current-carrying loop

\Sigma\vec{\tau}=2\times F\times (\frac{W}{2}\sin\phi)

=(ILB\sin\theta) (W\sin\phi)

=IAB\sin\phi

\vec{\tau}=\vec{\mu}\times\vec{B}

\vec{\mu}=IA\ \hat{n}

To calculate the net torque about the axis of rotation, consider the situation where the normal to the loop makes an angle with the magnetic field:

\phi

where

\implies