1. Coulomb's Law states that 'attractive' force between two charges is

F=-\frac{kQ_1Q_2}{r^2}

where

k=9.0 \times 10^9 N m^2/C^2

at least in air

F = -\frac{\left(9.0 \times 10^9 N m^2/C^2\right ) \left ( -8 \times 10^{-5} C \right) \left ( -8 \times 10^{-5} C \right ) }{\left (.25 m \right )^2} = -9.216 \times 10^2

A negative 'attractive' force is a repulsive force which is expected for like charges

2.) The force on a charge in an electric field is F=QE

F = \left ( 3.86 \times 10^{-5} C \right ) \left ( 1.75 \times 10^4 N / C \right ) = .6755 N

3.) R = \frac{ p L}{A} = \frac{\left ( 1.72 \times 10^{-6} \Omega cm \right ) \left ( 134 m \right ) }{2.07 \times 10^{-2} cm^2} \frac{100cm}{m} = 1.11 \Omega

(Also for what it is worth, 20 gauge wire has a cross sectional area of 5.18 \times 10^{-3} cm^2 )

4.) Energy = power x time

E = P t = \left ( 100 W \right ) \left ( 1 \mbox{wk} \right ) \frac{7 \mbox{days}}{ 1 \mbox{wk}} \frac{24 h}{1 \mbox{day}} = 16800 W h = 16.8 \mbox{kW}h

Cost = energy x rate

C = ER = \left ( 16.8 \mbox{kW}h \right ) \left ( \textdollar .20 / \mbox{kW} h \right ) = \textdollar 3.36

5.) Power = current x voltage

P = I V

The devices are in parallel, so the voltage drop across each device is 120 V

I=\frac{P}{V} = \frac{1000 W}{120 V} = 8.33 A

I=\frac{P}{V} = \frac{40 W}{120 V} = 0.33 A

I=\frac{P}{V} = \frac{550 W}{120 V} = 4.58 A

Voltage = Current x Resistance

V = I R

R = \frac{V}{I}

R = \frac{120 V}{8.33 A} = 14.4 \Omega

R = \frac{120 V}{.33 A} = 360 \Omega

R = \frac{120 V}{4.58 A} = 26.2 \Omega

6.)

W = F d \cos \theta

Since magnetic force is always perpendicular to velocity, it is always perpendicular to displacement so \cos \theta = 0

It also does not increase the speed - just changes the direction, so the kinetic energy of the particle remains unchanged - so there is no work done on the particle.

7.)

If you integrate the magnetic field around a loop, it equals the enclosed current times a constant…

The field around a wire is circular and has constant strength, so the integral is just 2 \pi r B

so we have

2 \pi r B = \mu_0 I

so

B = \frac{ \mu_0 I}{2 \pi r} = \frac{ \left ( 1.257 \times 10^{-6} T m/A \right ) \left ( 7.5 A \right ) }{ 2 \pi \left ( .500 m \right ) } = 3.00 \times 10 ^ {-6} T

8.) For a solenoid, we can use

B = \frac{\mu_0 N I}{L}

solving for I gives

I = \frac{ B L }{\mu_0 N} = \frac{\left ( 1.5 \times 10^{-3} T \right ) \left ( .150 m \right ) }{\left ( 1.257 \times 10^{-6} T m/A \right ) \left ( 750 \right ) } = .239 A

9.) A simple generator uses a gasoline engine to rotate magnets by coils of wire inducing a current in the wire. This induced current creates a counter magnetic field which provides the resistance to the gas engine.

10.) An electric motor usually produces an alternating current in a coil of wire within a constant magnetic field that causes a force on the coil of wire that drives the motor. A generator starts with another form of engine (wind, gasoline, thermal) and uses that energy to rotate magnets by coils of wire to induce currents.