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Solutions to PHY2049 Exam 3 by Paul Avery and Charles Thorn, Exams of Calculus

AMET UniversityCalculus

The solutions to exam 3 for the phy2049 course, taught by paul avery and charles thorn, during the spring 2004 semester. The solutions cover various problems related to electric circuits, magnetic fields, and electromagnetism.

Typology: Exams

2012/2013

Uploaded on 02/23/2013

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Solution to Exam 3
Paul Avery, Charles Thorn
PHY2049, Spring 2004
April 9, 2004
1. A resistance Rand an inductance Lare hooked up to a battery. The current in the
circuit builds up to one half of its steady-state value in 6.0 s. Find the inductive time
constant τL≡L/R (in sec).
Solution: The current in this RL circuit varies according to i=i0(1 −e−t/τ ), where
τ=R/L is the time constant. We get e−t/τ =1/2, or τ=t/ ln 2, yielding τ=8.66
sec.
2. Assume the average value of the vertical component of Earth’s magnetic field is 63 µT
(downward) in some region, which has an area of 3.19 ×105km2, and calculate the
net magnetic flux (in webers) through the rest of Earth’s surface (the entire surface
excluding that region). Take the negative direction to be inward.
Solution: The total magnetic flux must be zero, so the compensating flux must be
B×A=+2.0×107.
3. A generator with an adjustable frequency of oscillation is connected in series to a resis-
tance R, a capacitor of C=5.0F, and an inductor of inductance L. The amplitude of
the current produced in the circuit by the generator is maximized when the generator’s
frequency is 1.4 kHz. What is L(in henries)?
Solution: The amplitude is maximized at resonance, where f=1/2π√LC.Solving
yields L=2.6×10−9H.
4. Two straight conducting rails form a right angle where their ends are joined. A con-
ducting bar in contact with the rails starts at the vertex at time t=0andmoveswith
a constant velocity valong them, as shown in the figure. A magnetic field Bpoints
out of the page. If we write the emf as E=atn,whereaand nare constants, what is
the value of n?
1
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Solution to Exam 3

Paul Avery, Charles Thorn PHY2049, Spring 2004 April 9, 2004

  1. A resistance R and an inductance L are hooked up to a battery. The current in the circuit builds up to one half of its steady-state value in 6.0 s. Find the inductive time constant τL ≡ L/R (in sec). Solution: The current in this RL circuit varies according to i = i 0 (1 − e−t/τ^ ), where τ = R/L is the time constant. We get e−t/τ^ = 1/2, or τ = t/ ln 2, yielding τ = 8. 66 sec.
  2. Assume the average value of the vertical component of Earth’s magnetic field is 63 μT (downward) in some region, which has an area of 3. 19 × 105 km^2 , and calculate the net magnetic flux (in webers) through the rest of Earth’s surface (the entire surface excluding that region). Take the negative direction to be inward. Solution: The total magnetic flux must be zero, so the compensating flux must be B × A = +2. 0 × 107.
  3. A generator with an adjustable frequency of oscillation is connected in series to a resis- tance R, a capacitor of C = 5. 0 F , and an inductor of inductance L. The amplitude of the current produced in the circuit by the generator is maximized when the generator’s frequency is 1.4 kHz. What is L (in henries)? Solution: The amplitude is maximized at resonance, where f = 1/ 2 π

LC. Solving yields L = 2. 6 × 10 −^9 H.

  1. Two straight conducting rails form a right angle where their ends are joined. A con- ducting bar in contact with the rails starts at the vertex at time t = 0 and moves with a constant velocity v along them, as shown in the figure. A magnetic field B points out of the page. If we write the emf as E = atn, where a and n are constants, what is the value of n?

Solution: The emf E is related to the rate of change of magentic flux by dE/dt = −dφB /dt, where φB is the magnetic flux. In this case, the B field is constant but the area is changing. If we denote t as the time elapsed from when the bar crossed the vertex, then we see that the area is proportional to t^2 , thus the emf varies like E ∝ t. So n = 1.

  1. Suppose that a parallel-plate capacitor has circular plates with radius R = 35 mm and a plate separation of 4.1 mm. Suppose also that a sinusoidal potential difference with a maximum value of 130 V and a frequency of 60 Hz is applied across the plates; that is, V = 130 sin(2π 60 t). Find Bmax(R), the maximum value of the induced magnetic field that occurs at r = R. Solution: ∫ The B field can be calculated from the rate of change of electric flux using B · dl = μ 0 iD , where iD is the displacement current iD =  0 dφE /dt. Here φE is the electric flux. We are given the voltage V , from which the electric field E = V /d (d is the plate separation) can easily be calculated. The rate of change of the electric flux is dE/dt = 130(120π) cos(120πt)/d. When r = R, the integral is

∫ B·dl = B(2πR). Thus the maximum B field is Bmax =^ μ 0  0 130(120π)πR^2 /d/^2 πR^ or^ Bmax = 7800πR/dc^2 , where we used μ 0  0 = 1/c^2. Substituting yields Bmax = 2. 3 × 10 −^12 T.

  1. An ac generator emf is E = Em sin ωdt, with Em = 23.6 V and ωd = 296 rad/s. It is connected to a 16.5 H inductor. When the current is a maximum, what is the emf of the generator (in volts)? Solution: The current lags the emf by 90 degrees (a quarter cycle), so the emf is zero at that time.
  2. A beam of partially polarized light can be considered to be a mixture of polarized and unpolarized light. Suppose we send such a beam through a polarizing filter and then rotate the filter through 360 degrees keeping it perpendicular to the beam. If the transmitted intensity varies by a factor of 4.3 during the rotation, what fraction of the intensity of the original beam is associated with the beam’s polarized light?. Solution: Let the intensities of the unpolarized and polarized light be 1 and x, re- spectively. After going through the filter the intensity of the unpolarized light is 1/ 2 and the intensity of the polarized light will vary from 0 to x. Thus the ratio of the maximum to minimum intensity of the total beam is (1/2 + x)/(1/2) = 4.3. Solving yields x = 1.65. The fraction of polarized light is therefore 1. 65 / 2 .65 = 0.62.
  3. Consider the arrangement shown in the figure. A conducting rod, PQ, is lying on a U-shaped conducting wire, making good electrical contact with it, and both are supported on a nonconducting table. There is a uniform magnetic field pointing into the table. If the magnetic field strength is increased, which way does the rod move?