1. __________
2. __________
3. __________
4. __________
CQ ________
Total __________

(A nice improvement over the previous exam!)

Again, pace yourself . Don’t spend too much time on one problem. Write what you know and go on to another problem. If you have time, you can come back and think more about a problem. I can’t give partial credit for a blank page.

The centripetal force is supplied by the electric force from Coulomb's Law

(b) What is the effective current associated with this orbiting electron?

Current is given by

What is the period of this electron in orbit?

That is, an electron, with Q = e = 1.6 x 10 ^{- 19} C of charge passes by every 1.2 x 10 ^{- 16} s for a current of

2. Calculate the resistance at 20^oC of a 40-m, length of silver wire having a cross-sectional area of 0.40 mm².
The resistivity of silver is 1.59 x 10 ^{- 8} -m

R = r L /A
A = 0.4 mm² [ 1 m / 1000 mm ]² = 4 x 10 ^{- 7} m²
R = (1.59 x 10 ^{- 8} -m)(40 m)/(4 x 10 ^{- 7} m²)
R = 1.59

Reminder:

Conversion from mm to m seems to be easy but this conversion from mm² to m² caused more confusion than I think it should have. Since the mm in the area is squared the conversion fractor from mm² to m² must also be squared.

Don't forget to square the conversion factor in cases like this -- in area conversions.

3. Find the equivalent resistance in the circuit shown below for R₁ = 100 , R₂ = 500 , R₃ = 1000 , R₄ = 400 . If these are connected to a 12&endash;volt battery, what is the power supplied by the battery? Find the current through resistor R₁ and the power supplied to R₁.

Resistors R₂ and R₃ are in parallel so we will replace them with a single equivalent resistor, R_eq1;

1/R_eq1 = 1/R₂ + 1/R₃
1/R_eq1 = 1/500 + 1/1000
1/R_eq1 = (0.002 + 0.001)(1/)
1/R_eq1 = 0.003 (1/)
R_eq1 = (1/0.003) /
R_eq1 = 333

Now the three resistors, R₁, R_eq1, and R₄ are in series, so we replace them with a single equivalent resistor R_eq.

R_eq = R₁ + R_eq1 + R₄
R_eq = 100 + 333 + 400
R_eq = 833

How much power does the battery supply?

Remember this is the total power supplied by the battery.

How much current flows out of the battery?

I = V/R
I = 12 V / 833
I = 0.0144 A
I = 14.4 mA

The current that flows out of the battery is the same as the current that flows through resistor R₁ (and R₄).

What is the power absorbed by R_1?

P = I V = I² R = V² / R
P = I² R
P₁ = (0.014 4 A)² (100 )
P₁ = 0.021 W

Note or caution:

If you decided to use P = I V, be sure that you use the voltage across resistor R₁. And that voltage is not 12 V! However, if you did not realize that and went ahead and -- somehow -- did use 12 volts, look at what happens!

And that is the same as the total power supplied by the battery!.

Or, if you decided to use P = V²/R, be sure hat you use the voltage across resistor R₁. And that voltage is not 12 V! However, if you did not realize that and went ahead and -- somehow -- did use 12 volts, look at what happens!

and that is more -- considerably more! -- power than the total power supplied by the battery! Always ask if an answer is "reasonable", or "plausible" -- or possible! Always!

4. Consider the following RC circuit with R = 2 M and C = 5 mf and V_bat = 12 volts. Initially, the capacitor is uncharged. The switch is closed at time t = 0.

(a) What is the initial current immediately after the switch is closed?
(b) If we wait a very long time, what is the final charge that eventually collects on the capacitor?
(c) What is the “time constant” of this RC circuit?
(d) How much charge will there be on the capacitor 5 seconds after the switch is closed?
&emdash;&emdash;&emdash;&emdash;&emdash;&emdash;&emdash;&emdash;&emdash;&emdash;&emdash;&emdash;&emdash;&emdash;&emdash;&emdash;&emdash;&emdash;&emdash;&emdash;&emdash;&emdash;&emdash;&emdash;&emdash;&emdash;&emdash;&emdash;
(a) I _o = V/R = 12 V / 2 x 10⁶ W = 6 x 10 ^{- 6} A
(b) C = Q/V; Q_f = CV = (5 x10 ^{- 6} f) (12 V) = 60 x 10 ^{- 6} C = 60 mC
(c) t = RC = (2 x 10⁶ W)(5 x1 0 ^{- 6} f) = 10 s
(d) q = q(t) = Q_f e ^{- t/t} = [60 mC] e ^{- 5s/10s} = [60 mC] e ^{- 0.5}

Use P = I V or I = P / V

For the same voltage (110 V), the current is proportional to the power. So the 100 W bulb carries four times the current of the 25 W bulb.


Q2. How would you connect resistors so that the equivalent resistance is smaller than the individual resistance?

Resistors in paralled will provide an equivalent resistance that is smaller than the smallest individual resistance.


Q3. At a given instant, a proton moves in the positive x cirection in a region where there is a magnetic field in the negative z direction. What is the direction of the magnetic force? Does the proton continue to move in the positive x direction?

The vector v x B points along the positive y-direction. For a proton, q is positive, so the Force vector F = q v x B also points along the positive y-direction. With a force on it in the positive y-direction, the proton will be moved from its initial path. The proton will move in a circle in a plane parallel to the xy-plane.

Q4. Explain why two parallel wires carrying currents in the same direction attract each other.

Look at the magnetic field caused by the left current. At the location of the right current, B points down. The current in the conductor on the right is pointing into the page (or the screen) so the magnetic force F is to the left, toward the other current-carrying conductor. By Newton’s Third Law, the force on the left current will be just the opposite. These are forces of attraction.