PHYsics 1161
Hour Exam #2

March 3, 2003

You may have one sheet of equations (one side of an ordinary 8.5 x 11 inch sheet of paper).

1. An electron moves in a circular path, perpendicular to a magnetic field of B = 1.25 T, with a a speed of 1.2 x 104 m/s. What is the radius of the electron’s circular path?

Alternatively, we could have solved it symbolically first, as here,

2. A rectangular loop has dimensions of 2.0 cm (0.02 m) by 3.0 cm (0.03 m) and is initially between the poles of a large magnet, where the magnetic field strength is B = 0.050 T. The loop is oriented so the normal to the plane of the coil lies parallel to the direction of the magnetic field. The coil is removed from that region to a region with no magnetic field in 0.15 seconds. What is the average induced emf?

3. An electric light bulb, connected to an ordinary 120 V AC wall outlet, uses 60 W of power. What are
a) the rms current through the light bulb?
b) the maximum and minimum instantaneous values of the current?
c) the maximum and minimum instantaneous values of the power?
d) the resistance of the light bulb?

4. Consider a series RCL circuit with R = 80 W, C = 250 µF, L = 250 mH connected to a 12 V, 120 Hz AC source.
a), b) What are the reactances XC and XL?
c) What is the total impedance?
d) What current flows through the circuit?
e) What is the resonance frequency of this circuit?

5. Concept Questions:

i. A loop of wire is perpendicular to a magnetic field as shown in the sketch below. The magnetic field is initially directed out of the page with a field strength of 0.050 T. Over 2.9 s, the magnetic field decreases, vanishes, and then begins to increase, now directed into the page until it reaches a value of 0.040 T

What is the direction of the induced current in the loop of wire?
clockwise?
counterclockwise?
We began with magnetic flux pointing out of the page and ended with flux pointing into the page. The induced magnetic field will try to restore the change in the flux -- or restore the flux to its initial condition. That means the induced magnetic field will point out of the page. The induced electric current that causes this induced magnetic field must be counterclockwise, according to the right hand rule.

The minus sign simply says the direction of the induced emf is such that the corresponding induced magnetic field will oppose the change in the magnetic flux.

iii. Will the current in an AC series RC circuit increase or decrease when the frequency of the AC source increases? Explain why.
The current through a capacitor in a DC circuit is zero. A DC circuit has a frequency of zero. As the frequency increases, some charge will accumulate on the capacitor and then the direction of the voltage and current will change, the charge will decrease, go to zero, and then charge of the opposite sign will accumulate. As the frequency increases, the current increases.
Mathematically, we can see this from

XC = 1/( 2 f C)

As f increases, XC decreases. XC, the “capacitave reactance” is much like a resistance. A decrease in XC means an increase in the current I.

Concept Questions (cont’d)
Consider the following circuits. The three batteries are identical. The five light bulbs are identical. The power absorbed by a light bulb is directly related to its brightness.

iv. Rank the five light bulbs in order of their brightness (or power). Some bulbs may be equally bright.

A = D = E > B = C

The same voltage that is across A is also acros D and E. So the brightness -- or the power or the current -- of all three of these must be the same.
The voltage across B is the same as the voltage across C so the brighteness of B and C must be the same. The voltage across both B and C in series is the same as the voltage across A so the current through B and C will be less than the current through A; the brightness of B or C will be less than the brightness of B.

v. Rank the current through the three batteries.
This asks for the current through the batteries so it is a question of 1, 2, 3 or I1, I2, and I3. It is not a questions to be answered with A, B, C, D, and E.
Bulbs D and E have the same voltage across each or either of them as the voltage across A. So they each have the same current as that through bulb A. That means the current through battery 3 must be twice the current through battery 1. That is,

I3 > I1

The equivalent resistance for battery 2 is twice the resistance for battery 1 so the current supplied by battery 2 will be one half that supplied by battery 1. That is,

I1 > I2

Putting these together, we have

I3 > I1 > I2

vi. Bonus Question
Citroen DS21 was a
French automobile.
(Not much “Physics” in this one. It was meant only as a fun question for students who were in class on Friday.