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MASSACHUSETTS INSTITUTE OF TECHNOLOGY Department of Physics

8.02 Spring 2012

SOLUTIONS: Exam Two Equation Sheet

Please Remove this Tear Sheet from Your Exam Force Law:

( )ext extq F E v B

Force on Current Carrying Wire:

ext

wire

Id F s B

Source Equations:

32source source

( )ˆ( ) e e

dq dqk k

r

r r

E r rr r

2

3

ˆ( )

4

( )( )

4

o

source

o

source

Id

r

Id

s rB r

s r rB r

r r

r̂ points from source to field point Gauss’s Law:

closedsurface

enc

o

qd

E A

Gauss’s Law for Dielectrics

,

closedsurface

free enc

o

qd

E A

Electric Potential Difference:

B

AB B AA

V V V E ds

source

( ) ( ) e

dqV V k

rr r

V E

Potential Energy: U qV qV K 0

Capacitance:

C Q

V

221 1

2 2

QU C V

C

Capacitors in Parallel: C

eq C

1 C

2

Capacitors in Series:

1

Ceq

1

C1

1

C2

Current Density and Current:

open surface

I d J a

Ohm’s Law: V I R

where is the conductivityc c J E

where is the resistivityr r E J

Constants

ke

1

40

9 109 N m2 C2

0 4 107 T m A1

2

Problem 1 (25 points) In this problem you are asked to answer 5 questions, each worth 5 points. Question 1 (5 points) Wire 1 on the left below has a current I directed out of the page, as shown in the diagram. Wire 2 on the right has a current I directed into the page. In what direction does the magnetic field point at position P ?

The correct answer is __________C______________.

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Question 2 (5 points)

A parallel plate capacitor, initially with vacuum between the plates, is connected to a battery and charged to Q on the upper plate and Q on the lower plate respectively. The battery is left connected. A slab of material with dielectric constant is then inserted between the plates. a) The charge stored in the capacitor increases and the energy stored increases.

b) The charge stored in the capacitor remains the same and the energy stored increases.

c) The charge stored in the capacitor decreases and the energy stored increases.

d) The charge stored in the capacitor increases and the energy stored remains the same.

e) The charge stored in the capacitor remains the same and the energy stored remains the same.

f) The charge stored in the capacitor decreases and the energy stored remains the same.

g) The charge stored in the capacitor increases and the energy stored decreases.

h) The charge stored in the capacitor remains the same and the energy stored decreases.

i) The charge stored in the capacitor decreases and the energy stored decreases.

The correct answer is _________A_______________.

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Question 3 (5 points) Two resistors are made out of the same material, but have different dimensions, as shown in the figures. The current through these two resistors is in the directions shown. If the resistor on the left has a resistance of 1.00 ohm, the resistor on the right will have a resistance of

a) 1.50 ohm b) 2.00 ohm

c) 0.50 ohm

d) 0.75 ohm

e) 1.00 ohm

f) 0.67 ohm

g) None of the above.

The correct answer is ___________A_____________.

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Question 4 (5 points)

A conducting wire is attached to an initially charged spherical conducting shell of radius a (shown on the right in the figure above). The other end of the wire is attached to the outer surface of a neutral conducting spherical shell of radius 2a (shown on the left in the figure above) that is located a very large distance away (it is effectively at infinity). When electrostatic equilibrium is reached, the charge on the shell of radius 2a is equal to

a) half the final charge on the shell of radius a . b) twice the final charge on the shell of radius a .

c) four times the final charge on the shell of radius a . d) one-fourth the final charge on the shell of radius a .

e) None of the above.

The correct answer is __________B______________.

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Question 5 (5 points)

Consider a triangular loop of wire with three legs. The loop carries a current I in the directions shown. The magnetic field at the point P a) is the superposition of the vector sum of non-zero contributions from all three legs. b) is due only to leg 1. c) is due only to leg 2. d) is due only to leg 3. e) is the superposition of the vector sum of non-zero contributions from leg 2 and leg 3. f) is the superposition of the vector sum of non-zero contributions from leg 1 and leg 3. g) is the superposition of the vector sum of non-zero contributions from leg 1 and leg 2. h) is zero because all three legs have non-zero contributions but vectorially add up to

zero. i) is zero because each leg separately makes a zero contribution. The correct answer is __________F______________.

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Problem 2 (25 points) NOTE: YOU MUST SHOW WORK in order to get any credit for this problem. A dielectric rectangular slab has length s , width w , thickness d , and dielectric constant . The slab is inserted on the right hand side of a parallel-plate capacitor consisting of two conducting plates of width w , length L , and thickness d . The left hand side of the capacitor of length L s is empty. The capacitor is charged up such that the left hand side has surface charge densities L on the facing surfaces of the top and bottom plates respectively and the right hand side has surface charge densities R on the facing surfaces of the top and bottom plates respectively. The total charge on the entire top and bottom plates is Q and Q respectively. The charging battery is then removed from the circuit. Neglect all edge effects.

a) Find an expression for the magnitude of the electric field EL on the left hand side in terms of L , R , , s , w , L , 0 , and d as needed.

Using Gauss’s Law LL

o

E

b) Find an expression for the magnitude of the electric field ER on the right hand side in terms of L , R , , s , w , L , 0 , and d as needed.

Using Gauss’s Law for dielectrics RR

o

E

c) Find an expression that relates the surface charge densities L and R in terms of

, s , w , L , 0 , and d as needed.

The potential difference on the left side is LL

o

dE d

. On the right hand side it is

RR

o

dE d

. Since these must be equal we must have RL

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d) What is the total charge Q on the entire top plate? Express your answer in terms of L , R , , s , w , L , 0 , and d as needed.

L RQ L s w sw

e) What is the capacitance of this system? Express your answer in terms of , s ,

w , L , 0 , and d as needed.

The potential difference is LL

o

dE d

, so the capacitance is

L R o oR

L L

o

L s w sw w wQC L s s L s s C

dV d d

f) Suppose the dielectric is removed. What is the change in the stored potential

energy of the capacitor? Express your answer in terms of Q , , s , w , L , 0 , and d as needed.

Since the battery has been removed, the charge on the capacitor does not change when we do this. So the change in the energy stored is

2 2 2

2 2

1 1 1 1change in energy

2 2 2

1 1

2 2

o o

o oo

Q Q Q

C C C C

Q d d Q d

Lw w L L s sw L s s

2 2 1change in energy

2 2o o

L s s L sQ d Q d

w wL L s s L L s s

Problem 3 (25 points) NOTE: YOU MUST SHOW WORK in order to get any credit for this problem. A positively charged ion with charge q and mass m , initially at rest, is accelerated across a gap by an electric potential difference with magnitude V . The ion enters a region of uniform magnetic field B

1 pointing into the page of the figure below and

follows a semi-circular trajectory of radius R1. The ion is again accelerated across the

gap, increasing its speed, by an electric potential difference that has the opposite sign but

has the same magnitude V . The ion then enters a region of uniform magnetic field

B2 pointing into the page of the figure below and follows a semi-circular trajectory of

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radius R2 2R

1. What is the ratio B

2/ B

1 of the magnitudes of the magnetic field in the

different regions?

2v v

In general we have v vm qB m

q B R RR m qB

In region 1 the speed is given by 21 1

21v v

2

q Vm q V

m

Thus we have that 11

1 1

2v q Vm mR

qB qB m

And in region 2 the speed is given by 22 2

41v 2 v

2

q Vm q V

m

Thus we have that 22

2 2

4v q Vm mR

qB qB m

Since R2 2R

1, we therefore have

2 1 2 1

4 2 1 22

q V q Vm m

qB m qB m B B

So 1 2

2 1

12 or

2

B B

B B

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Problem 4 (25 points) NOTE: YOU MUST SHOW WORK in order to get any credit for this problem. A current loop, shown in the figure below, consists of two arc segments, both with a common center and the same opening angle , and two straight line segments joining the arc segments. The inner arc segment has radius R and the outer arc segment has radius

R a . The current I1 through the loop is counterclockwise. Point P is the common

center of the two arc segments.

a) What is the direction and magnitude of the magnetic field B

at the point P ?

The sides will not contribute because the cross product is zero there.

1 11

outer arc 2 22

ˆˆ ˆ( ) 2

4 24 4o o oI II d

ds R ar R a R a

s r

B r z z

1 11

outer arc 22

ˆˆ ˆ( )

4 44o o oI II d

dsr R aR a

s r

B r z z

1 11inner arc 2 2

ˆˆ ˆ( )

4 4 4o o oI II d

dsr R R

s rB r z z

Therefore the total field at P is given by

1 1 1

1

1 1ˆ ˆ ˆ( )

4 4 4

ˆ( )4

o o o

o

I I I

R a R R a R

I a

R R a

B r z z z

B r z

Two fixed conducting rails are arranged as shown in the figure below. A metal bar of

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length s is placed at the origin, initially held in place, and a current I2 runs through

the rails and bar clockwise. The bar is then released. You may assume that the length of the bar is very short and that the magnetic field you calculated in part a) is uniform over the length of the bar. You may neglect the magnetic field due to the current through the rails.

b) What is the direction and magnitude of the magnetic force acting on bar the instant it is released?

The field at point P is inward from our calculations above, and the current in the bar is

downward, so the total force on the bar is in the positive x direction, with

1

2ˆ4

oI aI s

R R a

F x