Midterm Exam - Week 4 also has a midterm exam.SHOW ALL WORK. I can only give partial credit based on shown work should the work you submit have any validity to solving the problem, EVEN if you did not choose the correct answer. Obviously, non multiple choice questions must show work and not just an answer as well.See ATP for a special post I made about Kirchoff's Laws. You will learn how to solve electric loop circuits using the two rules I review. I have worked out two examples for you to review. Do this BEFORE you try to solve the Kirchoff's Law problem on the mid term exam!!

1.
A beam of light traveling in air is incident on a transparent plastic at an angle of incidence of 50o.
The angle of refraction is 35o. What is the index of refraction of the plastic?
A) 1.56
B) 1.34
C) 1.67
D) 1.43
Solution:


n2= 1.3355
2.
A long, straight wire carries a current of 2.5 A. Find the magnitude of the magnetic field 25 cm from
the wire.
Solution:


3.
If a current is flowing with a value of 5.9 A, how much electron charge passes any single point
in 25 seconds?
A) 2.3 x 1020 e
B) 1.9 x 1020 e
C) 9.2 x 1020 e
D) 1.5 x 1020 e
Solution:
No of electron = 5.9*25/e=9.206 E+20
4.
See the figure. The net force on the 1 nC charge is zero. What is q?

A) zero
B) 0.68 nC
C) 4.0 nC
D) 1.5 nC
Solution:

Total force due to two 2nC charges is twice the vertical component of the individual force.

For Net force to be zero. Verical component by q should be equal to the above force and should act in
opposite direction.



5.
Estimate the average power output of the Sun, given that about 1350 W/m2 reaches the upper
atmosphere of the Earth. The distance from the Sun to the Earth is 1.5 × 1011 m.
A) 4 × 1026 W
B) 3 × 1026 W
C) 2 × 1026 W
D) 1 × 1026 W
Solution:

6.
Which of the equations here is valid for the circuit shown?

A) 2 - 2I1 - 2I2 - 4I3 = 0
B) 6 - I1 - 2I2 = 0
C) -2 - I1 + 4 - 2I2 = 0
D) 4 - I1 + 4I3 = 0
Solution:
Applying KVL in left loop
We get
-2 - I1 + 4 - 2I2 = 0
7.
Starting from rest, a proton falls through a potential difference of 2700 V. What speed does it acquire?
A) 3.6 × 105 m/s
B) 2.4 × 105 m/s
C) 7.2 × 105 m/s
D) 4.8 × 105 m/s
Solution:
Potential difference = work done per unit charge
Therefore work done = potential difference * charge
Charge on a proton = 1.6 * 10^-19 C
Potential difference = 2700 V
Therefore
work W = 2700* 1.6 * 10^-19 J = 4.33 * 10^-16 J
Let speed of proton = v
Initial kinetic energy = 0 because the proton is initially at rest.
Final kinetic energy = 1/2 mv^2
where m = mass of proton
Change in K.E. = 1/2 mv^2 - 0 = 1/2 mv^2
By work energy theorem
change in K.E. = work done
Or 1/2 mv^2 = W
Or v = sqrt(2W/m)------...
Or v = sqrt{2 * 4.33 * 10^-16/(1.67 * 10^-27)}
= sqrt{(2 * 4.33/1.67) * 10^(-16 + 27)}
= 7.2 * 10^5 m/s
8.
If the force between two charges increases by a factor of 16 because the charges are moved closer
together, how much closer are they moved?
A) 4
B) 2
C) 8
D) 16
Solution:

Means Charges is moved 4 times.
9.
A concave mirror with a radius of 30 cm creates a real image 40 cm from the mirror.
What is the object distance?
A) 24 cm
B) 70 cm
C) 5.0 cm
D) 10 cm
Solution:



u= 24cm
10.
A flux of 4.0 × 10-5Wb is maintained through a coil for 0.50 s. What emf is induced in this coil by this flux?
A) 2.0 x 10-5 V
B) No emf is induced in this coil.
C) 8.0 x 10-5 V
D) 4.0 x 10-5 V
Solution: