Exam Solutions for PHY2049 Spring 2008 - Problem 1 to 10, Exams of Calculus

The solutions to exam 3 of the phy2049 spring 2008 course. The problems cover various topics in physics, including electromagnetism, circuits, and capacitors. Students can use these solutions to check their understanding of the exam material.

Typology: Exams

2012/2013

Uploaded on 02/23/2013

dron.singh
dron.singh 🇮🇳

4.3

(10)

79 documents

1 / 5

Toggle sidebar

This page cannot be seen from the preview

Don't miss anything!

bg1
PHY2049 Spring 2008 Exam 3 Solutions
1
Prof. Darin Acosta
Prof. Selman Hershfield
April 9, 2008
Exam 3 Solutions
1. A metal rod is forced to move with constant velocity of 60 cm/s [or 90 cm/s] along
two parallel metal rails, which are connected by metal at one end (see figure). The rails
are separated by 20 cm, and there is a magnetic field of 2 T coming out of the page. If
the rod has resistance 50 Ohms and the rails and connector have negligible resistance,
what force is required to keep the rod moving at constant velocity?
(1) 2 x 10-3 N (2) 1 x 10-3 N (3) 3 x 10-3 N (4) 4 x 10-3 N (5) 5 x 10-3 N
The induced EMF (magnitude) is given by Faraday’s Law
()
|| B
dd
B
Lx vt BLv
dt dt
ε
Φ
== +=
The current is given by
i
R
ε
=
This gives rise to a force on the rod carrying this current (recall Ch.28):
F
iLB=
Plugging in gives F = 2 x 10-3 N for v=60 cm/s, and F = 3 x 10-3 N for v=90 cm/s
2. The current in the large loop is in the clockwise [or counterclockwise] direction and
increasing. What is the direction of the induced current in loop 2 and in loop 3,
respectively?
(1) clockwise, clockwise
(2) clockwise, counterclockwise
(3) counterclockwise, clockwise
(4) counterclockwise, counterclockwise
(5) no induced current
pf3
pf4
pf5

Partial preview of the text

Download Exam Solutions for PHY2049 Spring 2008 - Problem 1 to 10 and more Exams Calculus in PDF only on Docsity!

Prof. Darin Acosta Prof. Selman Hershfield April 9, 2008

Exam 3 Solutions

  1. A metal rod is forced to move with constant velocity of 60 cm/s [or 90 cm/s] along two parallel metal rails, which are connected by metal at one end (see figure). The rails are separated by 20 cm, and there is a magnetic field of 2 T coming out of the page. If the rod has resistance 50 Ohms and the rails and connector have negligible resistance, what force is required to keep the rod moving at constant velocity?

(1) 2 x 10 -3^ N (2) 1 x 10 -3^ N (3) 3 x 10 -3^ N (4) 4 x 10 -3^ N (5) 5 x 10 -3^ N

The induced EMF (magnitude) is given by Faraday’s Law

| | B ( )

d d BL x vt BLv dt dt

The current is given by

i R

This gives rise to a force on the rod carrying this current (recall Ch.28): F = iLB

Plugging in gives F = 2 x 10 -3^ N for v=60 cm/s, and F = 3 x 10-3^ N for v=90 cm/s

  1. The current in the large loop is in the clockwise [or counterclockwise] direction and increasing. What is the direction of the induced current in loop 2 and in loop 3, respectively? (1) clockwise, clockwise (2) clockwise, counterclockwise (3) counterclockwise, clockwise (4) counterclockwise, counterclockwise (5) no induced current

For a clockwise current, the field points into the page within ring 1 and out of the page outside of ring 1. For example, the field lines look like below (use the right-hand rule):

Thus, since the current is increasing, the field strength is increasing, which means in the vicinity of rind 2 and 3 the field is growing out of the page. This induces and EMF and current in rings 2 and 3 which must oppose this increase in magnetic flux (Lenz’s Law), thus those rings also have clockwise currents induced which create fields pointing into the page inside the loops (thus subtracting from the increasing magnetic flux from ring 1).

  1. An alternating source drives a series RLC circuit with an EMF amplitude of 5V. The current lags the EMF by 45º. When the potential difference across the inductor, vL, reaches its maximum positive value of 6V [or 4V], what is the potential difference across the capacitor, vC , including sign? The potential difference across the resistor is 0. (1) -2.5V (2) -0.5V (3) 1V (4) -1V (5) 3.5V

This is alternating current and EMF, not DC! Recall that for AC circuits:

( ) sin ( ) sin

m m

t t i t i t

The potential difference across the inductor is C cos( )

di v L L t dt

This is a maximum when ω t − φ = 0, 2 π,... Let’s take 0, then the EMF is:

( ) sin sin sin 5sin 45 3.5V 2

ε t = ε m ω t = ε m ω t − φ + φ = ε m φ = = =

Then by Kirchoff’s loop rule: 0 3.5 0 6 2.5V for 6 3.5 0 4 0.5V for 4

R L C C L C L

v v v v v v v

z

x

  1. A parallel-plate capacitor with circular plates of radius 0.05 m is being charged by a constant current of 0.5 A. What is the magnitude of the magnetic field in μT at a radius of 0.025 m from the central axis connecting the centers of the plates?

(1) 1 (2) 2 (3) 3 (4) 4 (5) 0

The displacement current between the capacitor plates equals the real current going into the capacitor. It is distributed uniformly between the plates since the electric field there is uniform. Thus the fraction of the current enclosed by a circular path of radius r is 2 disp encl , (^2)

r i i R

Thus by the Ampere-Maxwell Law 2 (^0 )

0 6 2

1 10 T

r B ds rB i R ir B R

⇒ = = ×

  1. A paramagnetic [or diamagnetic] material is placed in an external magnetic field Bext. The magnitude of the magnetic field inside the material is:

(1) > |Bext| (paramagnetic) (2) < |Bext| (diamagnetic) (3) = |Bext| (4) exactly zero (5) cannot be determined from information given

Recall that the material magnetization goes as: BM = μ 0 M = χ Bext. For paramagnetic

substances χ >0 and for diamagnetic ones, χ <0. So the total field Btot = Bext + BM is

larger for paramagnetic substances and smaller for diamagnetic ones.

  1. The average intensity of light from an incandescent light bulb is 300 mW/m^2 [or 600] on a particular surface. Assuming that the light is in the form of an electromagnetic plane wave, what is the maximum magnetic field amplitude, B (^) m?

(1) 5 x 10 -8^ T (2) 7 x 10 -8^ T (3) 3.5 x 10 -8^ T (4) 21 T (5) 15 T

0 0

2 0 8 8

since

0.3 [or 0.6] 2 5 10 T (or 7 10 T)

av m m rms rms

m m

av m

m

I S E B E B

E

c B c I S B

B

− −

⇒ = × ×

  1. A beam of initially unpolarized light is sent into a stack of 4 polarizing sheets, where the polarizing direction of each sheet is rotated +60º with respect to the previous sheet. What fraction of the incident intensity is transmitted by the stack of 4 sheets?

(1) 1/128 (2) 1/256 (3) 1/64 (4) 1/16 (5) 27/

Unpolarized light passing through one Polaroid reduces the intensity by 1/2. After that, each Polaroid reduces the intensity by a factor cos^2 θ. So the over reduction is:

3 2 3 2 0

cos 60 2 2 2 128

I

I

⋅ = ⎛^ ⎞^ = ⎛^ ⎞ = ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠

o

  1. Light traveling horizontally enters a right prism as shown in the figure. The index of refraction of the prism is n=1.6. At what angle is the light deflected from horizontal?

(1) 31º (2) 26º (3) 19º (4) 45º (5) 0º

There are 2 refractions from the 2 surfaces. And since the second surface is not parallel to the first, the light does not return to the incident direction.

1 1 2 2

sin 45 1.6sin

1.6sin 45 sin

31

θ θ θ θ

⇒ = o

θ 1

θ 2

45-θ 1 45-θ 1