Download 1 (-4.32x10 + 3.2x10 ) Newtons

AP Physics
3rd Quarter Test Review
KEY
k = 8.99 x 109 Nm2/C2
milli = 10-3
ε0 = 8.85 x 10-12 C2/Nm2
micro = 10-6
nano = 10-9
e = 1.6 x 10-19 C
pico = 10-12
Mega =10+6
Electrostatics
1.
Charge, Field, and Potential
a.
A proton is placed in a uniform electric field that has a value of (3.00i – 2.50j) Newtons/Coulomb. What is the
force on the proton?
F = qE = (1.6 x 10-19 C) (3.00i – 2.50j) = (4.8 x 10-19i – 4.0 x 10-19j) Newtons
b.
An electron is placed at (3.00, 2.00, -1.50) m. The electric field in that area is given by the function E =
(3.00(N/Cm2)x2i – 2.50(N/Cm3)y3k) Newtons/Coulomb. What is the force on the particle?
E=3(3)2i – 2.5(2)3k = (27i – 20k) N/C
F = qE = (-1.6 x 10-19 C) (27i – 20k) N/C = (-4.32 x 10-18i + 3.2 x 10-18k) Newtons
c.
A dipole has a moment of 6.70 x 10-3 j Coulomb-meters. It is placed in a uniform field
Newtons/Coulomb. Find the initial force, torque, & potential energy on the dipole.
ΣF = 0
τ = p x E = (6.7 x 10-3j) x (3.00i) = -0.0201 k Nm
U = - p•E = -(6.7 x 10-3j) • (-2.5j) = 0.0168 J
d.
Indicate on the diagram below where the electric field is strongest and where it is the weakest.
(3.00i – 2.50j)
What is the difference between electric field lines and
electric field vectors?
Field lines start at the charge and use density to show
magnitude.
Vectors start at the point where field is evaluated & use
length to show magnitude.
weak
Draw field lines for the non-conducting sheet of charge
below.
strong
+
+
+
+
+
+
+
+
_
e.
Calculate the initial acceleration of the electron in #1b.
a=
∑ F = (-4.32x10
m
-18
i + 3.2x10-18 k ) Newtons
= (-4.742x1012 i + 3.513x1012k )m/s 2
-31
9.11x10 kg
1
f.
A proton is initially traveling at a speed of 3.0x105 m/s when it enters a uniform field of 5.00 V/m. What is the final
speed of the proton if it travels 3.0 mm through this field?
∆V = Ed = (5V / m)(0.003m) = 0.015V
W = q∆V = (1.6 x10 −19 C )(0.015V ) = 2.4 x10 −21 J
W = ∆K =
1
m ( v 2f − vi2 )
2
2 ( 2.4 x10 −21 J ) + (1.67 x10−27 kg )(3 x105 m / s )2
2W + mvi2
=
= 300, 005m / s
vf =
m
(1.67 x10−27 kg )
g.
Electric potentials are drawn for an electric field in the diagram below. Is the field uniform? yes or no (circle one)
Explain your answer. The equipotential surfaces are evenly spaced.
What is the direction of the field? left
A
h.
100 V
0V
150 V 200 V
B
250 V
300 V
350 V
If a 10.0 C charge has 30.0 kJ of kinetic energy at point A, how much kinetic energy will the 10.0 C charge have
when it reaches point B.
W = ∆K
K f = W + K i = q∆V + K i = −(10C )(100V ) + 30, 000 J = 29, 400 J
i.
If the function for electric field is given by 3.0x2 + 2.0x – 5.0, find a function for the potential at any point in the
field.
V = − ∫ E • dx = ∫ 3 x 2 + 2 x − 5dx = − x 3 − x 2 + 5 x + C
j.
If the function for potential is given by 5.0xy + 3.0z2, find the electric field strength at (2.0, 3.0 -1.0) m.
∂V
= −5 yi = −15i
∂x
∂V
Ey = −
= −5 xj = −10 j
∂y
∂V
Ez = −
= −6 zk = 6k
∂z
E = (−15i − 10 j + 6k )V / m
Ex = −
2
2.
Coulomb’s Law and Field and Potential of Point Charges
a.
Find the force on the bottom right charge. Each charge is -3.0 C.
q
q
2.0 m
P
q
q
4.0 m
2
4
θ = 26.6
tan θ =
F2 =
F3 =
F4 =
k (3C ) 2
= 2.02 x1010 N , 270
(2m) 2
k (3C ) 2
(
(4m)2 + (2m) 2
= 4.05 x109 N , −26.6
k (3C ) 2
= 5.06 x109 N , 0
(4m) 2
∑ F = 2.37 x10
b.
)
2
8
N , −68.5 = (8.68 x109 i − 2.22 x1010 j) N
Point P is located at the center of the rectangle above. Determine the potential at point P due to all 4 charges.
n
V =∑
i =1
kqi
4k (−3C )
=
= −4.8 x1010 V
2
2
ri
(1m) + (2m)
c.
How much work must be applied to bring a 5.0 Coulomb charge from infinity to point P in the rectangle above?
Wapplied = q∆V = (5 C)(-4.8 x 1010V – 0V) = -2.4 x 1011 J
d.
Calculate the total electrical potential energy of the five charge system, after the 5.0 Coulomb charge is placed at
point P.
kq1q2
r
1
1
1
1 1
1
U f = ∆U + U i = −2.4 x1011 J + k (3C )2  +
+ +
+ + 
20 2
20 2 4 
4
∆U = Wapplied =- 2.4 x 1011 J or you could add up each combination of
U=
2 
3
U f = −2.4 x1011 J + k (3C ) 2  +
= −8.37 x1010 J

20 
2
3
3.
Fields and Potentials of Other Charge Distributions
a.
Find the potential at point P (directly above the left end of the line of charge) due to the uniform line of charge.
The line of charge has a total charge of 7.0 C and a length of 3.0 m. Its thickness is negligible.
P
z=0.5 m
r
θ
x=0
x=L
dx
kdq
r
dq Q
=
dx L
Q
dq = dx
L
kQ dx
V=
L ∫ r
V =∫
r = x2 + z 2
kQ x = L dx
L ∫x =0 x 2 + z 2
x
tan θ =
z
x = z tan θ
V=
dx = z sec2 θ dθ
x2 + z 2
z
kQ x = L z sec2 θ dθ
kQ x = L z sec2 θ dθ
kQ x = L sec 2 θ dθ kQ x = L
V=
=
=
=
sec θ dθ
L ∫x =0 z 2 tan 2 θ + z 2
L ∫x =0 z 2 (tan 2 θ + 1) L ∫x = 0 sec θ
L ∫x =0
sec θ =
V=
kQ x = L
 sec θ + tan θ
sec dθ 
∫
L x =0
 sec θ + tan θ
2
 kQ x = L  sec θ + sec θ tan θ
=

∫ 
 L x =0  sec θ + tan θ

 dθ

u = sec θ + tan θ
du = (sec 2 θ + sec θ tan θ )dθ
kQ x = L du kQ
V=
=
ln sec θ + tan θ
L ∫x =0 u
L
x= L
x=0
kQ
=
ln
L
x2 + z 2 x
+
z
z
x=L
=
x =0
x=L
kQ 
ln( x 2 + z 2 + x) − ln z 
 x =0
L 
kQ 
kQ 
V=
ln( L2 + z 2 + L) − ln z − ln z + ln z  =
ln( L2 + z 2 + L) − ln z 




L
L
k (7C ) 
V=
ln( (3m) 2 + (0.5m) 2 + (3m)) − ln(0.5m)  = 5.2 x1010 V


(3m)
4
b.
Find the electric field and potential at point P that is 0.50 m above a ring of charge. Point P lies on the central axis
of the ring of charge. The total charge of the ring is 1.0 microCoulombs. The radius of the ring is 0.10 m.
dEsinΘ
dEcosΘ
dE
Θ
P
E = k∫
cos θ =
E=
Θ
z
E=
r
dq
k
= 2
dq cos θ
2
r
R + z2 ∫
z
R2 + z2
k
z
kz
Qkz
dq
=
dq =
3 ∫
3
2 ∫
2
2
R +z
R +z
( R2 + z2 )2
( R2 + z 2 )2
2
(1µ C )k (0.5m)
( (0.1m)
V = k∫
dq
=
r
2
+ (0.5m)
kq
R +z
2
2
3
2 2
)
=
= 33,905V / m
k (1µ C )
(0.1m) 2 + (0.5m) 2
= 17, 630V
R
R
5
c.P Find the electric field and potential at point P that is 0.5 m above a uniform disk of charge. Point P lies on the
central axis of the disk of charge. The total charge of the ring is 5.0 microCoulombs. The radius of the ring is
0.1m.
Q
(5µ C )
=
= 1.59 x10 −4 C / m2
2
π r π (0.1m) 2
dq = σ dA = σ 2π rdr
σ=
z
zdq
dE =
3
2 2
4πε 0 ( z + R )
2
E=
dr
σz
4ε 0
from electric field for single ring
4πε 0 ( z + R )
2rdr
R
0
3
2 2
2
3
(z2 + r2 )2
u = z2 + r2
r
du = 2rdr
R
E=
σz
4ε 0
E=
V=
∫
zσ 2π rdr
=
∫
r =R
−
du
r =0
(u )
=
3
2
1
2
r =R
σz u
4ε 0 −1/ 2
=
r =0

1.59 x10−4 C / m 2 
1−

2ε 0


−σ z 
1
1 σ 
z
− =
1−
 2


2ε 0  z + R 2 z  2ε 0 
z 2 + R2 
0.5m
( 0.5m ) + ( 0.1m )
2
dq
1
σ dA
σ
2π R ' dR '
σ
=
=
=
∫
∫
∫
4πε 0 r
4πε 0
r
4πε 0
r
4πε 0
1
∫
R '= R
R '= 0
2

 = 1.75 x105 V / m


2π R ' dR '
z 2 + R '2
u = z 2 + R '2
du = 2 R ' dR '

 1


σ R '= R du ' σ  1   − 2 +1
=
V=
u


4ε 0 ∫R '= 0 u 4ε 0  − 1 + 1 
 2 
V=
σ
2ε 0
(
)
z 2 + R2 − z =
R '= R
σ 12
=
u
2ε 0
R '= R
R '= 0
σ z 2 + R '2
2ε 0
R
0
R '= 0
1.59 x10−4 C / m 2
2ε 0
(
)
(0.5m) 2 + (0.1m) 2 − (0.5m) = 8.9 x104 V
d.
Draw field lines for a set of parallel plates.
e.
Draw field lines for a long uniformly charged wire or thin cylindrical shell. Use a cross-sectional view.
+
+
+
+
+
+
-
+
6
f.
Draw field lines for a thin spherical shell. Use a cross-sectional view.
+
4. Gauss’ Law
a.
A non-uniform electric field given by E = (1.2x)i + (2.5y2 +y)j + (3.0z3)k pierces the Gaussian cube shown below.
The back bottom left vertex of the cube is located at the origin. The cube has a side length of 1.0 m. Determine
the electric flux through each face and the total flux through the cube.
y
A = 1m 2
Φ = E•A
Φ front = 3(13 )k • 1k = +3.0 Nm 2 / C
Φ back = 0
x Φ right = 1.2(1)i • 1i = +1.2 Nm 2 / C
Φ left = 0
Φ top = (2.5(12 ) + 1) j • 1j = +3.5 Nm 2 / C
Φ bottom = 0
∑ Φ = +7.7 Nm
2
/C
z
b.
Begin with Gauss’ law and derive an expression for the electric field at a point a distance of r from a uniformly
charged infinitely long thin wire. The wire has a linear charge density of λ. Include a relevant sketch.
ε 0 ∫ E • dA = qenclosed
ε 0 E 2π rl = λl
λ
E=
2πε 0 r
7
c.
The diagram below shows a spherical shell made of porcelain, i.e., it is non-conducting. Charge is uniformly
distributed throughout the shell. Find an expression for the electric field for r<a, for a<r<b, and for r>b. Make a
plot of r vs. b. Assume the total charge on the shell is Q.
for r<a
E
ε 0 ∫ E • dA = qenclosed = 0
E=0
a
for a<r<b
ε 0 ∫ E • dA = qenclosed
b
ρ=
a
b
r
Q
4
π (b3 − a 3 )
3
Q
4
π (r 3 − a 3 )
4
π (b3 − a 3 ) 3
3
3
Q (r − a 3 )
E=
4πε 0 r 2 (b3 − a 3 )
ε 0 E 4π r 2 =
for r>b
ε 0 ∫ E • dA = qenclosed
ε 0 E 4π r 2 = Q
E=
d.
Q
4πε 0 r 2
Find the surface charge density of the conducting sheet of charge below. Assume that the sheet is very large
compared to the small -3.0 microCoulomb charge suspended with a non-conducting string. The -3.0
microCoulomb charge has a mass of 3.0 milligrams. Assume everything is at rest.
+
+
+
+
+
+
+
+
+
+
+
+
∑F
y
30°
T
T cos θ = mg
mg
T=
cos θ
∑ Fx = 0
T sin θ = qE
qσ
mg tan θ =
qE
mg
=0
σ=
σ=
ε0
ε 0 mg tan θ
q
ε 0 (3 x10−6 kg ) g tan 30
= 50 pC / m 2
(3µ C )
Note: 1 mg = 10-6 kg
8
Conductors, Capacitors, and Dielectrics
5.
Electrostatics with Conductors
a.
Give at least two possible explanations for why there is no charge inside a conductor.
Charge redistributes to make like charges far away from each other.
If there was charge on the inside, there would be field. This would cause the charges to be in perpetual motion.
b.
Why is a conductor an equipotential surface?
Otherwise, charge would constantly be in motion on the surface.
c.
In the diagram below, the sphere on the left, A, initially has a radius of 0.05 mm and a charge of 3.0 nC. The
sphere on the right, B has a radius of 0.03 mm and a charge of 50.0 nC. The spheres are 2.0 meters apart. Find
the net charge on each sphere, after they are connected with a conducting wire.
VA = VB
A
B
kq A kqB
=
rA
rB
qA
qB
−
=0
.05mm .03mm
q A + qB = 53nC
1
.05
1
−1
qA
0
=
.03
q
53
1 B
q A 33nC
=
qB 20nC
d.
A 5.0 pF charge is brought near a large non-conducting sheet of charge that has a charge density of 5.0 µC/m2.
If the 5.0 pF charge has a mass of 3.0 µg, calculate its acceleration.
qσ
∑ F = ma = qE = 2ε
a=
0
qσ
(5 pF )(5µ C / m 2 )
=
= 471m / s 2
2mε 0
2(3 x10 −9 kg )ε 0
Note: 1µ g = 1x10 −9 kg
e.
On the graph below, make a sketch of V vs. r for a conducting sphere with a radius R and charge Q. Label
important points.
for r<R: V=kQ/R
for r>R: V=kQ/r
V
kQ/R
r
R
9
6.
Capacitors
a.
A parallel plate capacitor has an area A = 2.00 x 10-4 m2 and a plate separation d = 1.00 mm. Find its
capacitance.
C=
b.
ε0 A
d
=
ε 0 (2 x10−4 m 2 )
(1x10−3 m)
= 1.77 pF
If the capacitor above is then connected to a 12.0 V battery, what is the charge on the capacitor?
q = CV = (1.77 pF )(12V ) = 21.2 pC
c.
How much energy is stored in the above capacitor when it is fully charged?
1
1
U = CV 2 = (1.77 pF )(12V )2 = 0.127 nJ
2
2
d.
The battery is disconnected and a 0.8 mm slab of porcelain is inserted between the plates of the capacitor.
Calculate the electric field in the slab of porcelain.
E=
e.
Calculate the electric field in the air gaps between the capacitor plates.
E=
f.
σ
q
21.2 pC
=
=
= 1846V / m
κε 0 κ Aε 0 6.5(2 x10−4 m2 )ε 0
σ
= κ (1846V / m) = 12, 000V / m
ε0
Calculate the potential difference across the capacitor.
V = Ed = (1846V / m)(0.8 x10 −3 m) + (12, 000V / m)(0.2 x10−3 m) = 3.88V
g.
Calculate the new capacitance.
c=
h.
q 21.2 pC
=
= 5.5 pF
V
3.88V
If the battery had not been disconnected, the charge stored on the capacitor would have changed. What would
the new charge be?
q = CV = (5.5 pF )(12V ) = 66 pC
i.
Calculate the work applied when the dielectric was inserted.
1
Wapplied = ∆U = (5.5 pF )(12V ) 2 − 0.127 nJ = .269nJ
2
j.
The outer radius of a cylindrical capacitor is increased while the capacitor is connected to a battery. How will the
charge change?
C = 2πε 0
L
and q=CV
ln(b / a )
b↑C↓q↓
10
k.
The outer radius of a spherical capacitor is increased while the capacitor is connected to a battery. How will the
charge change?
ab
a
=4πε 0
and q=CV
b−a
1− a / b
b↑C↓q↓
C = 4πε 0
i.
Find the equivalent capacitance of the circuit below. C1 = 3.0 µF, C2 =4.0 µF, and C3 = 5.0 µF.
Ceq =
j.
C1C2
(3µ F )(4 µ F )
+ C3 =
+ 5µ F = 6.71µ F
C1 + C2
3µ F + 4 µ F
In the circuit below, the resistor R = 6000Ω , the emf is 12 V, and the capacitor C = 3.0 nF. Find the current in the
circuit when the switch is first placed at a.
i=
V
12V
=
= 0.002 A
R 6000Ω
k.
Find the current in the circuit above, a long time after the switch is in position a. i=0
l.
What is the maximum charge on the capacitor above?
q = CV = (3nF )(12V ) = 36nC
11
m. How long does it take for the capacitor above to reach half its maximum charge?
Q = CV (1 − e− t / RC )
1
CV = CV (1 − e− t / RC )
2
1
= 1 − e −t / RC
2
1
− = e −t / RC
2
ln 2 = t / RC
t = RC ln 2 = ln 2(6000Ω)(3nF ) = 1.2 x10 −5 sec
n.
After a long time, the switch is flipped to position b. Calculate the initial current.
i=-0.002A (opposite direction of the initial current when the switch is flipped to a).
o.
What is the current a long time after the switch is flipped to position b. i=0
p.
What is the time constant of the circuit?
RC = (6000Ω)(3nF ) = 1.8 x10 −5 sec
Electric Circuits
7. Current, Resistance, Power
a.
If there is a current of 5.0 A in a resistor, how much charge passes through the resistor in 2.0 hours?
∆q = i∆t = 2hr (
b.
5C 3600 s
)(
) = 36000C
s
hr
What is the potential difference across the above resistor, if it has a resistance of 250.0 Ω?
V = iR = (5 A)(250Ω) = 1250V
c.
A conductor of a uniform radius 1.2 cm carries a current of 3.0 A produced by an electric field of 120 V/m. What is
the resistivity of the material?
E
E
Eπ r 2 (120V / m)π ( 0.012m )
ρ= =
=
=
= 1.81x10 −2 Ωm
J i/ A
i
3A
2
d.
Determine the resistance of a cylindrical resistor at 300K that is made of silver and has a radius of 5.00 mm. The
resistor is 1.0 cm long.
from p. 669, ρ 0 = 1.62 x10−8 Ωm and α =4.1x10-3 K −1
ρ = ρ 0 + ρ0α (T − T0 ) = 1.62 x10−8 Ωm + (1.62 x10 −8 Ωm)(4.1x10-3 K −1 )(300 K − 293K ) = 1.67 x10−8 Ωm
R=
ρL
A
=
(1.67 x10−8 Ωm)(0.01m)
= 2.1x10−6 Ω
2
π (0.005m)
12
e.
An electric heater is constructed by applying a potential difference of 110 V to a nichrome wire of total resistance
8.00 Ω. Find the current carried by the wire and the power rating of the heater.
V 2 (110V ) 2
P=
=
= 1513W
R
8Ω
V 110V
i= =
= 13.75 A
R
8Ω
f.
On the circuit below, indicate how you would connect an ammeter and a voltmeter to read the current and voltage
respectively, in R2.
A
i3
i1
V
i2
g.
Given the values of E1 = 12.0 V, E2 = 24.0 V, E3 = 9.0 V, R1 = 6.0 Ω, and R2 = 2.0 Ω, determine the current in
each branch of the circuit.
i1 + i3 = i2
i1 − i2 + i3 = 0
ξ1 − i1R1 − i2 R2 − ξ 2 − i1 R1 = 0
12 − 12i1 − 2i2 − 24 = 0
−12i1 − 2i2 = 12
ξ 2 + i2 R2 + i3 R1 − ξ3 + i3 R1 = 0
24 + 2i2 + 12i3 − 9 = 0
2i2 + 12i3 = −15
1
−1
−12 −2
0
i1
2
1 i1
0
0 i2 = 12
12 i3
−15
−.72
i2 = −1.69 amperes
i3
−0.97
13
h.
Find the equivalent resistance of the network shown below.
Req = R1 +
R2 R3 R4
1
1
= R1 +
= R1 +
1
1
1
R3 R4 + R2 R4 + R2 R3
R3 R4 + R2 R4 + R2 R3
+ +
R2 R3 R4
R2 R3 R4
i.
Explain why the resistance of an ammeter must be low.
If it was high, it would drastically reduce the circuit current since ammeters are connected in series.
k.
Explain why the resistance of a voltmeter must be high.
If it was low, more of the current would be diverted from the tested branch. This would alter the value of the
potential drop.
14
6.17Ω
iC 1
2
iE
iD
iB
6.73Ω
3
1
iF
3.73Ω
2
3
3.17Ω
l.
In the circuit above, R1 = 1.0 Ω, R2 = 2.0 Ω, R3 = 3.0 Ω, R4 = 4.0 Ω, R5 = 5.0 Ω, R6 = 6.0 Ω, R7 = 7.0 Ω.
R8 = 8.0 Ω, E1 = 10.0 V, and E2 = 10.0 V. Find the value i1 and i2.
loop 1: 10V – 9iD – 10 V = 0
iD=0
to calculate i1, determine the equivalent resistance for all resistors to the left of E1 (see diagram)
i1 =
10V
= 1.62 A
6.17Ω
junction 1: i1 = iC + iD
iC=1.62A
loop 2:
6iB + 3 iC-10V=0
iB = (10-3(1.62))/6=0.86A
junction2: iE= iC- iB=1.62A-0.86A=0.76A
loop3: 8i2 = 7iF
junction 3:
iE = i2 + iF
0.76A = i2+(8/7)i2
i2=0.36A
15
R
r
m. A battery has an emf of 15.0 V. The terminal voltage of the battery is 11.6 V when it is delivering 20.0 W of power
to an external load resistor R. What is the value of R?
V2
R
V 2 (11.6V ) 2
R=
=
= 6.73Ω
P
20W
P=
n.
In the above problem, what is the internal resistance of the battery?
V 11.6V
=
= 1.72 A
R 6.73Ω
V = ξ − ir
i=
11.6V = 15V − (1.72 A)r
r = 1.97Ω
Magnetostatics
8.
Forces on Moving Charges in Magnetic Fields
a.
Explain why a magnetic field cannot do work.
Magnetic fields cannot do work because the force they exert is always perpendicular to the displacement.
b.
An electron moving along the positive x axis perpendicular to a magnetic field experiences a magnetic deflection
in the negative y direction. What is the direction of the magnetic field? -z
c.
An electron is projected into a uniform magnetic field B = (1.4i + 2.1j) T. Find the vector expression for the force
on the electron when its velocity v – 3.7 x 105 j m/s.
F=qvxB = (-1.6x10-19C)(-3.7x105jm/s x 1.4i) = -8.29x10-14 k Newtons
d.
A proton is moving in a circular orbit of radius 14 cm in a uniform magnetic field of magnitude 0.35 T directed
perpendicular to the velocity of the proton. Find the orbital speed of the proton.
∑ F = ma
v2
qvB = m
r
qBr (1.6 x1019 C )(0.35T )(0.14m)
v=
=
= 4.69 x106 m / s
m
(1.67 x10 −27 kg )
16
e.
A crossed field velocity selector has a magnetic field of magnitude 0.0100 T. What electric field strength is
required if 10.0 keV electrons are to pass through undeflected?
K=
1 2
mv
2
 1.6 x10−19 J
(10, 000eV ) 
eV

v = 5.93 x107 m / s
 1
−31
2
 = (9.11x10 kg )v
 2
FE = FB
qE = qvB
E = vB = (5.93 x107 m / s )(0.01T ) = 5.93 x105V / m
9.
Forces on Current Carrying Wires in Magnetic Fields
a.
A wire having a mass per unit length of 0.50 g/cm carries a 2.0 A current horizontally to the south. What are the
direction and magnitude of the minimum magnetic field needed to lift this wire vertically upward?
West
0.5 g  1kg   100cm  0.05kg


=
cm  1000 g   1m 
m
FB = mg
ilB = mg
B=
b.
mg 0.05kg (9.8m / s 2 )
=
= 0.245T
li
m
2A
A current of 17.0 mA is maintained in a single circular loop of 2.00 m circumference. A uniform magnetic field of
0.800 T is directed parallel to the plane of the loop. Calculate the magnetic moment of the loop.
π d = 2.0m
d = 2.0m / π = 0.64m
µ = NiA = (1)(17 x10−3 A)
c.
π (0.64m) 2
4
= 5.4 x10−3 Am2
What is the magnitude of the torque exerted on the loop in the above problem by the magnetic field when the
plane of the loop is parallel to the magnetic field?
τ = µ xB = (5.4 x10−3 Am 2 )(0.8T ) = 4.3x10−3 TAm 2
17