PHY2054 -- Exam 2 -- 11 July 2011

SOLUTIONS

The class average on this exam was 11.3/18 — slighty above 60% .  Good!

Scores on this 18-point exam will be normalized to a 25-point basis when final averages are calculated.

Note: Two questions (numbers 3 and 12) were discarded.

See explanations at each item number.

Item (question) numbers below refer to the master exam. Click Here for the master exam. 

Correct answer is (1) unless solution differs (see # 1, for example)

[In the solutions, RHR = Right Hand Rule]

Also,  it appears that many web browsers aren't compatible with Greek letters:

Translation for equations:  m = 'mu', F = 'phi', p = 'pi' (note different font)

1. The vertical lightning bolt acts as a momentary straight current.

Thus B = m₀ I/2pr , where r = 100 m

Induced emf E = -N dF/dt . Here, dF/dt = A dB/dt so E= NA[m₀ I/2pr] dI/dt

and N = E/A [2pr/m₀ dI/dt . dI/dt = 10 e^6 A/(10e^-6s) = e^12 A/s .

Thus N = 30V/(5e^-4 m²) [(2p)(100m)/( 4pe^-7)(e^12) = 30 turns

The B field lines around the vertical bolt are horizontal (parallel to Earth's surface). 

In order for them to penetrate the coil, its plane must be perpendicular to Earth's surface.

2. For charge q moving perpendicular to B, r = mv/qB .

r is diminishing,so B must be increasing.

The force is directed toward the center of the spiral so

by the RHR, q must be negative.

3. There was an error here, missed in proofreading. The coil should have been a solenoid where L is proportional to N².

But flat coils don't quite fit the solenoid model! Therefore this question is deleted from the exam.

4. The impedance of the series RLC circuit Z = …[R² + (X_L - X_C)²] . At resonance, X_L= X_C

so Z = R ; thus voltage across R = applied voltage.

5. When S is closed, the current path is through both R₁ and R₂ ; the back emf in L momentarily blocks current through L.

Hence I = V_o/(R₁ + R₂) = 45/(10 + 5) = 3A

6. After a long time, the (now) steady current through L is limited only by R₁,

so I = 45/10 = 4.5A

7. U = Ω LI² = Ω (0.3) (4.5)² = 3J

8. The electron moves parallel to B so no magnetic force.

9. F/l = m₀ I₁I₂/2pd = m₀/2p (I²/d) . F'/l = m₀/2p(4I²)/3d = 4/3 m₀/2p (I²/d) = 4F/3

10. If t is large, there is no back emf so V(L) = 0; thus V(R) = V(battery), so 0, 20V

11. For 8:1 step-down and secondary current 25A, the primary current must be 25/8.

P = VI so e^4 W = 25/8 V_P ==> e^4 W = 25/8V_P ==> V_P = e^4 (8/25) = 3200 V

12. This question was inadvertently based on material not covered by this exam.

It is therefore deleted.

13. In charging C, the series resistance controls the charging current,

thus t = RC = (e^4)(2 e^4) = 2s

14. When S is opened, C discharges through both resistors, so

t = RC = (2 e^4)(2 e^4) = 4s

15. if the secondary resistance is 2W , I_S = 24V/2W = 12A

Therefore I_P = 12A/5 = 2.4A

16. With the switch open, the two lamps are connected across 12+ 8 = 20V.

Since they are identical their resistances are the same, so each has 20/2 = 10 volts across it. 

But when the switch is closed, A has 12 V across it and B has 8 V.

So A is now brighter, B dimmer.

17. By the RHR, the top wire is forced upward, out of the page, and the bottom wire

downward. There is no force on the side wires. So the torque is about the x-axis.

18. B due to straight current = m₀ I/2pr , downward.

B' due to current loop = m₀ I/2r, upward. So total B = B' - B

= m₀ I/2r - m₀ I/4pr = m₀ I/2r(1 - 1/2p) = [(4p e^-7) (3A)/2(2e^-2)](1 - 0.159)

= 7.9 e^-5 T.

19. E = dF/dt = A dB/dt

A = coil area = 1.81e^-3 m² ==> E = 1.81e^-3 m² (3T/0.4s) = 1.358 e^-3 V

I = E/R = 1.358 e^-3 V/1.5W = 9 e^-3 A .

Induced current must produce B opposing the increasing applied B (Linz's law!), so induced B is downward, hence [RHR] I is clockwise.

20. t = BIA =50T(10 e^-3)(p)(1m)² = 1.57 e^-2 Nm