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Unformatted text preview: Problem 1 (8 pts) An electron moves through a region “of uniform electric ﬁeld < 0, 250, 0> V/m and uniform magnetic ﬁeld < 0.2, 0.3, 0> T.
Calculate the net force on the electron at an instant where its velocity is < 0, 500, 0> m/s. Express your answer as a vector. A ,x A ~ .3. A —‘
TV 5 EFT.=?E +7.VX3
e”; /, rf/ée—zczxoﬂsoﬂ) + (142—17) (0500,07 «(J9.03.0) Problem 2 (10 pts) The diagram shows magnetic ﬁeld vectors on a cylindrical surface of height
11 cm and radius 7 cm. On the top surface of the cylinder, the magnetic ﬁeld
points upward and has magnitude Bl = 1.7 T. Everywhere on the side of the
cylinder, the magnetic ﬁeld makes an angle of 65 degrees with a line
perpendicular to the surface, and has a magnitude B; = 2.3 T. On the bottom
surface, the magnetic ﬁeld again points upward and has an unknown
magnitude Ba. Find the magnitude B3. Show all steps in your work. Eggnog/1225.3“ =0
gm, +3..?.A,+ 339/1320 31w+ Bar/neswmv —— 6w:— : [51an + 131(co549°)(9nrhl 33an
53: a, + 02; (comm.
: a. 7 7—) + W“) (“mam “0 (o. 07) ’33 c 775 7" Problem 3 (30 pts) You pull a rectangular loop of wire with an applied force F1 = 0.26 N at a constant speed v, with the left portion of the loop in
a region of uniform magnetic ﬁeld B directed into the page, as shown in Figure 1. You then rotate the loop 90 degrees and
pull it again through the same region of magnetic ﬁeld at the same speed v with a force F 2, as shown in ﬁgure 2. The length
of the loop is L = 34 cm and the width is w = 12 cm. Outside the region of magnetic ﬁeld, B = 0. Calculate the magnitude of the force F 2. Start from fundamentals and show all steps in your work. If you use an equation not given on the formula sheet, you must derive it.
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A bar made out of a conducting material has a rectangular cross section. It is 4 cm high, 3 cm deep, and 13 cm long. The bar
is connected to a battery whose emf is 2.7 volts by low—resistance connecting wires, and carries a steady current. A uniform
magnetic ﬁeld of 0.95 tesla is applied perpendicular to the bar, coming out of the page (using some coils that are not shown).
A voltmeter is connected across the bar and reads a steady potential difference of +0.043 mV (mV = 1e3V). The connections
across the bar were carefully placed directly across from each other to eliminate false readings corresponding to the much
larger potential difference along the bar. Remember that a voltmeter gives a positive reading if the lead labeled “+” is
connected to the higher potential location. 2.7 volts Voltrnetei + 13 cm (a, 2 pts) On the diagram above, draw the distribution of charges on the surface of the bar due to polarization of the bar by the
magnetic force. (b, 2 pts) On the diagram above draw and label an arrow representing E” , the electric ﬁeld due to the battery and surface charges, which drives the current in the bar. (c, 12 pts) Assume that there is only one kind of mobile charge in this material. Are the mobile charges in this material
positive (holes) or negative (electrons)? As an explanation of your reasoning, you must clearly draw and label on the
diagram above: 0 A mobile charge, with its sign clearly indicated. 0 An arrow representing the drift velocity of the mobile charge (of appropriate sign) in the bar 0 An arrow representing the magnetic force on a mobile charge in the bar. 0 An arrow representing the transverse (perpendicular) electric ﬁeld E, in the bar. c Hmées ARF MEGATNF (d, 10 pts) In the steady state, what is the drift speed of the mobile charges? Start from fundamentals. If you use an equation
not given on the formula sheet, you must derive it. Show your work and reasoning clearly. _ IN STE/WY STATE, lAV‘i: Elk v: E;_ Opal/3,“.
= F ”rs ' _
FM EJ— ma! 0,0‘133—3V 0'” T .. E: ———=
av/3'aé A h ‘qu v= Alisa—”5W: v: 5:!— 30,0014")? .
B m . (e: 8 pts) What is the mobility of the bar? v: a II (4'5—2
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£7": lbvlll — gﬂ:&70.87‘: u: 5,75e’5 V Problem 5 (18 pts) A block of metal (in static equilibrium) is positively charged.
An imaginary box, 16 cm long on one side, is embedded into the
metal so that half of its length is inside the metal and half is
outside. The box is 5 cm high and 6 cm wide. The charge enclosed by the box is 5e10 C. On the right surface
of the box, the electric ﬁeld due to the metal is perpendicular to
the surface and has a magnitude of 8000 V/m (it is
approximately uniform on this right surface). For each part below, be sure to explain your answer and/or show
all your work. (a, 4 pts) What is the electric ﬂux on the right surface of the
box? 3.2M: EA: Ewk
>_(gooo)(.0€)(.04): 07"] Vom (b, 4 pts) What IS the sum of the electric ﬂux on the surfaces of the left half of the box (the half that is inside the metal)?
g 13 E CAuSé E: 0 ws/EE MWAL A?" $7347". 64
) (/91, 10 pts) Find the sum of the electric ﬂux on the surfaces, excluding the right surface, on the half of the box that IS outside
of the metal. In other words, ﬁnd the sum of the electric ﬂux on the top, bottom, front, and back of the right half of the box
outside the metal FROM 6/1055' 5 LAW, £97? + {06th + ERBT : M @E Ind/Jr: (611mm: ___.——— HAL F 5‘92 FM? 6 o e; o (\‘1
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 Spring '08
 WEISENSTEIN
 Physics

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