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Descripción: cap 8 problemas
Ejemplo 28.1 Fuerzas entre dos protones en movimiento pág. 959
Dos proton protones es se mueven mueven paralelo paraleloss al eje x en sentidos opuestos (fgura 28.2) con la misma rapidez v (pequeña en comparación con la rapidez de la luz c). En el inst instan ante te que que se ilus ilustr tra a ca calc lcul ule e las las !uerz !uerzas as el"c el"ctr tric icas as # magn"ticas so$re el protón de la parte superior # determine la razón de sus magnitudes. %olución& 'a !uerza el"ctrica est dada por la le# l e# de oulom$. *ara encontrar la !uerza magn"tica primero de$emos determinar el campo magn"tico que produce el protón de la parte in!erior en la posición del de arri$a.
PLANTEAR: %e usa la ecuación (21.2) que e+presa la le# de oulom$. 'a ecuación (28.2) da el campo magn"tico magn"tico de$ido al protón protón in!erior # la le# de la !uerza magn"tica ecuación (2,.2) da la !uerza magn"tica resultante so$re el protón superior.
EJECT EJECTAR: De acuerdo con la le# de oulom$ la magnitud de la !uerza el"ctrica so$re el protón de arri$a es& 2
q Fe= K 2 r
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!alta
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Ejemplo 28.-& ampo magn"tico de un solo alam$re n conductor largo # recto conduce una corriente de 1./ 0. 0 qu" distancia del eje del conductor el campo magn"tico generado tiene una magnitud
−4
B =0.5 × 10 T
(apro+imadamente (apro+imadamente el campo magn"tico
terrestre en *itts$urg)3
!"LC#$N: 'as l4neas de campo de este conductor recto son c4rculos c4rculos cu#a dirección va a estar determinada por la regla de la mano dereca como lo muestra la fgura&
B=
%AT"!
•r =
μ0 I 2 πB
r53 ( 4 × 10− T 7
m
)( 1.0 A )
μ 0 I 2 πr
%e despeja 6r7 de la ecuación original dada # se sustitu#en los valores por los datos que dan
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¿ 4 × 10− m=4 mm 3
5 1./ 0 −7
μ0= 4 π × 10 T ∙
m A
*ro$lema 28.n electrón se mueve a /.1//c como se muestra en la fgura E28.-. alcule la magnitud # dirección del campo magn"tico que este electrón produce en los siguientes puntos cada uno situado a 2.00 μm
desde el electrón& a) puntos 0 # 9: $) punto : c) punto D.
!"LC#$N: ; El campo el"ctrico de$ido a una carga en movimiento es& B=
μ0 qv sin θ ∙ 2 4 π r
D0<=%& −7
μ0= 4 π × 10 T ∙
7
v = 3.0 × 10 m / s −6
r =2 × 10 m
m A
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a !e sustitu&en 'os va'ores en 'a e(ua(i)n respe(tiva: m −7 4 π × 10 T ∙ B=
−19
A ( 1.6 × 10 ∙
4 π
−8
B =6.00 × 10 T =60 nT
7
)( 3.0 × 10 m / s ) sin sin 30 C )( ( 2 × 10− m) 6
2
acia a!uera de la pgina en 0 # 9.
* En e' punto C −7
B =( 1 × 10 T ∙
−7
B =1.20 × 10 T
m )( 1.6 × 10−19 C )( )( 3.0 × 107 m / s )( 2 × 10−6 m )2 A
0!uera de la pgina.
( En e' punto % B5/< #a que sin(18/)5 /
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Ejemplo 28.>& ampo magn"tico de dos alam$r alam$res es 'a fgura 28.,a es la vista de los e+tremos de dos alam$res largos rectos # paralelos que son perpendiculares perpendiculares al plano +# cada uno de los cuales conduce una corriente pero en sentidos opuestos. a) alcule
B
en los puntos
P1 , P2 y P 3
.
$) Deduzca una e+presión e+presión para B en cualquier punto del eje + a la dereca del alam$re 2. Fig. +,-
%='?@&
Ana'izar: *or el principio de superposición el campo magn"tico en cada punto va a ser
B = B1 + B 2
. %e va a utilizar la ecuación
para para poder poder o$tene o$tenerr las magnitu magnitudes des de
B 1 y B 2
B=
μ 0 I 2 πr
de estos campos #
utilizar la regla de la mano dereca para determinar las direcciones correspondientes. 'a fgura muestra que B , B y B = Btotal en cualquier 1
2
punto: se de$e confrmar que las direcciones # magnitudes relativas mostradas son correctas.
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•
•
⃗ 1= B ⃗ 2= B
μ0 I
(
2 π 2 d
)
μ0 I
(
2 π 4 d
)
μ 0 I
=
=
4 πd
μ0 I 8 πd
Enton(es:
⃗ B total=⃗ B 1+ ⃗ B2 =
− μ I 0
J +
μ0 I
^
4 πd
J = ^
8 πd
− μ I 0
J (*unto P1 ) 8 πd ^
En el punto P a una distancia 6d7 entre los dos alam$res 2
B 1 y B 2
tienen am$os dirección en 6#7 positiva # los dos tienen la misma magnitud:
⃗ 1= B ⃗ 2= B
μ 0 I 2 πd
Enton(es:
⃗ total =⃗B1+ B ⃗2= B
En el punto
P3
μ 0 I
J +
μ 0 I
^
2 πd
J =
− μ I 0
^
2 πd
J ( Punto P2 ) ^
πd
la regla de la mano dereca indica que B est en 1
dirección 6#7 positiva # que B est en dirección 6#7 negativa. Este 2
punto se encuentra a -d del alam$re 1 # a una distancia d del
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* En cualquier punto so$re el eje 6+7 a la dereca del alam$re 2 B 1 y B 2
estn en las mismas direcciones que en P . Este punto est a una distancia 6+Ad7 del alam$re 1 # a una distancia 6+Bd7 del alam$re 2 entonces el campo total es& 3
⃗ B total=⃗ B1 + ⃗ B2 =
¿−
μ 0 Id
J ^
2
2
π ( x −d )
μ 0 I
( +d )
2 π x
J − ^
μ0 I
( −d )
2 π x
J ^
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Ejemplo 28.C& ampo magn"tico de una $o$ina na $o$ina con 1// espiras circulares con radio de /.C/ m conduce una corriente de ./ 0. a) alcule el campo magn"tico en un punto a lo largo del eje de la $o$ina a /.8/ m del centro. b) %o$re el eje a qu" distancia desde el centro de la $o$ina la magnitud del campo es 1 8
de la que tiene en el centro3
!"LC#$N: %AT"! @5 1// a . n e g i r o l e n e a r t s i g e r e s e u q d u t i n g a m a l e d e n e i t o p m a c l e e u q l e n e + e d m / C . / = r o l a v l e s e . + a d a n e d r o o c a l e d o d a d r o l a v n u n e s e 5 ./ 0
a= /.C/ m −7
μ0= 4 π × 10 T ∙
m A
μ 0∋ a
B x =
2
3
2
2
2
( x + a )
2
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0.80 m
¿
(
−7
μ 0= 4 π × 10 T ∙
B x =
m A
)
( 100 ) ( 5.0 A ) ( 0.60 m )
2
3
2
[( ¿ ¿ 2 +( 0.60 m) )] 2
2
−4
B x =1.1 × 10 T
*
onsiderando la ecuación usada anteriormente queremos encontrar un valor de 6+7 tal que&
1
1
3 2
2 2
( x + a )
1
= ∙ 8
3 2
2 2
(0 +a )
%e le da vuelta a las !racciones # se elevan am$os lados la potencia 2F-: entonces& x = √ 3 a = 1.04 m
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E/emp'o +,. Campo de un (ondu(tor 'argo0 re(to & portado portadorr de (orriente En la sección 28.C se o$tuvo la le# de 0mpGre empleando la ecuación (28.H) para el campo de un conductor largo recto # que transporta$a corriente. Ievierta este proceso # utilice la le# de 0mpGre para encontrar la magnitud y dirección de 1 en esta situación.
%olución #%ENT# #%ENT#F#C F#CAR: AR: Esta situación presenta simetr4a cil4ndrica por lo que se utiliza la le# de 0mpGre para encontrar el campo magn"tico en todos los puntos u$icados a una distancia r del conductor.
PLANTEAR: %e toma como tra#ectoria de integración un c4rculo con radio r centrado en el conductor # en un plano perpendicular a "ste como en la fgura 28.1Ca (sección 28.C). En cada punto 1 es tangente a este c4rculo. De acuerdo con la elección de la tra#ectoria de integración la le# de 0mpGre Jecuación (28.2/)K es&
∮ B dl= B (2 πr ) 2 # de inmediato se deduce la ecuación (28.H) 0 I
μ 3
#
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total que producen estas cargas en el punto * que se encuentra a 1.,nm de cada uno.
e d l a i r o t c e v a m u s a l n o c a r t n e u c n e e s l a t o t o c i t " n g a m o p m a c l e # s o c i t " n g a m s o p m a c n e c u d o r p
%='?@& B = Bal!a + Bele"tron =
D0<=%
μ0 v 4 πr
2
( e sin sin 40 # + 2 e sin140 # )
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m −7 −19 5 m C 2.50 × 10 4 π × 10 T ∙ 1.6 × 10 A
B=
(
)( )(
−9 4 π 1.75 × 10 m
(
s
2
)
) ( sin 40 # + 2 sin sin 140 140 # )
−3
B =2.52 × 10 T = 2.52 mT
*ro$lema 28.C Dos cargas puntuales positivas q =+ 8.00 μC # q $ =+ 3.00 μC se desplazan en relación con un o$servador en el punto * como se ilustra en la fgura. 'a distancia d es /.12/m v =4.50 × 10 m / s # 6
6
v $ =9.00 × 10 m / s
. a) uando las dos cargas estn en las u$icaciones
que se india en la fgura ules son la magnitud # la dirección del campo magn"tico neto que producen en el punto *3 $) ules son la magnitud # la dirección de las !uerzas el"ctrica # magn"tica que cada carga ejerce so$re so$re la otra3 ul es la razón entre la magnitud de la !uerza el"ctrica # la magnitud de la !uerza magn"tica3 c) %i la dirección de v $ se invierte de manera que las dos cargas se desplacen en la misma dirección ules son la magnitud # la dirección de las !uerzas magn"ticas que cada carga ejerce so$re la
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−6
6
q =+ 8.00 × 10 C v =4.50 × 10 m / s −6
6
q $ =+ 3.00 × 10 C v $ = 9.00 × 10 m / s
d5 /.12/m −7
μ0= 4 π × 10 T ∙
m A
En la parte a) r5d #
⃗r
es perpendicular a
casos por lo tanto& r5 2d
Enton(es: a
μ 0 qv q$ v $ ( 2 + 2 ) B total= B + B = 4 π d d $
0.120
¿ ¿ ¿2 ¿
( 8 × 10− C ) ( 4.50 × 10 m / s ) ¿ 6
6
m
⃗ v
en todos los
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F C =&
q1 q2 r
2
() 8 × 10− C ) ( 3 × 10− C ) 6
=( 9 × 10 % m / C 9
2
2
6
2
( 0.240 )
=3.75 %
La relación entre la fuerza magnética y la eléctrica es: F "
2
3.75 % C = = = 2.22 × 10 3 % −3 F B v1 v 2 1.69 × 10 %
: la !uerza el"ctrica es ms
grande que la magn"tica.
( 'a dirección de las !uerzas magn"ticas se invierten cuando se invierte la dirección de una sola velocidad sin em$argo la magnitud de las !uerza permanece igual.
*ro$lema 28.C, Dos alam$res largos recto # paralelos estn separados por una distania de 1.//m como muestra la fgura. El alam$re de la izquierda conduce una corriente I de C.// 0 acia el plano del papel. a) 1
ules de$en ser la magnitud # el sentido de la corriente I para 2
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B=
μ 0 I 2 πr
a I 1 =6.00 A
B 1 y B 2
de$en ser iguales #
opuestos para que el campo
1.00 m
resultante en 6*7 sea cero. B va 2
acia la dereca: entonces I va 2
I 2
acia !uera de la pgina
0.50 m
B 1 B2
*
%AT"! −7
μ = 4 π × 10 T ∙
m
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•
(
−7 4 π × 10 T
∙
2 π
(
• I 2=
*
0.50 m 1.50 m
B2
0.50 m • B1
)(
L
(
m A
)
∙
(
6.00 A 1.50 m
6.00 A
)
=
(
−7 4 π × 10 T
2 π
∙
m A
)
∙
(
I 2 0.50 m
)=2.00 A
B 1 • B1=
μ 0 I 1 2 π r1
)(
= 2 × 10− T ∙ m ∙ 7
A
6.00 A 0.50 m
)
=2.40 × 10− T 6
)
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1.//m
B 1 • B2=
/.8/m
• B 2=
(
μ 0 I 2 2 π r2
−7 2 × 10 T
∙
)(
m ∙ A
2.00 A 1.80 m
)=
−7 5 ' 00 × 10 T
I 2
B y B tienen ngulos rectos entre los dos entonces la magnitud se da 1
2
por& B =√ B 1 + B2 =√ ( 2 ' 00 × 10 T ) + ( 5 ' 00 × 10 T ) =2.06 × 10 T 2