Download - © 2001-2005 Shannon W. Helzer. All Rights Reserved. 1 Chapter 29 – Magnetic Fields Due to Current
© 2001-2005 Shannon W. Helzer. All Rights Reserved. 1
Chapter 29 – Magnetic Fields Dueto Current
© 2001-2005 Shannon W. Helzer. All Rights Reserved. 2
Biot Savart Law WS 11 #1 Whenever a current flows through a wire, a magnetic field is
induced about the wire. The direction of this magnetic field may be found by
applying the right hand rule to the cross product below. Point your thumb in the direction the current is flowing. The magnetic field wraps around the wire in the same
direction as your fingers wrap around the wire. When the current is going up (remember, current is in the
opposite direction as the electrons are moving), we find that the magnetic field around the wire is counterclockwise.
When the current is going down, we find that the magnetic field around the wire is clockwise.
We will determine the magnitude of the magnetic field on the next slide.
34o ids r
dBr
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Biot Savart Law – Magnetic Field of a long, Straight Wire WS 11 #4 In this problem we will determine the magnetic field generated at
point P by the upper half of a long current carrying wire. First determine the magnetic field direction. The vector ds is in the same direction as the current.
34o ids r
dBr
ds
3
sin
4o idsr
dBr
2
sin
4o ids
dBr
r
PR
s
sin sinR
r
1/ 22 2r R s
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Biot Savart Law – Magnetic Field of a long, Straight Wire WS 11 #4
ds
r
PR
s
1/ 2 21/ 22 2 2 2
1
4o R
dB idsR s R s
3/ 22 24o R
dB idsR s
2
sin
4o ids
dBr
1/ 22 2sin
R R
r R s
1/ 22 2r R s
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Biot Savart Law – Magnetic Field of a long, Straight Wire WS 11 #4
ds
r
PR
s
3/ 22 24o R
dB idsR s
3/ 20 2 24oiR ds
BR s
1/ 22 2 2
0
4oiR s
BR R s
1 04
oiBR
4oiBR
We have just found the magnetic field generated by the upper half of the wire.
In order to determine the magnetic field generated by the entire length of the wire at point P, simply multiply our result by two.
2oiBR
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Magnetic Field due to a Current Carrying Arc WS 11 #5 Determine the magnetic
field at point P due to an arc shaped current carrying wire.
Remember, must be in radians.
ds
r
2
sin
4o ids
dBr
24o ids
dBR
90 , r R
2 04oiB dsR
, ds r s rd Rd
2 04oiB dsR
2 04oiB RdR
4
oiBR
P
R
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Magnetic Field Example WS 11 #6 Determine the magnetic
field at point P due to current carrying wire shown.
The arc in the wire covers 3/4 radians.
This problem may be done in three segments.
What is the angle between R1 and ds1?
What is the angle between R3 and ds3?
We already derived the equation for the arc portion of the wire.
2
1
3
1
1 2
sin 00
4o
idsdB
r
2ds
3ds
1ds
1R
3R
3
3 2
sin 1800
4o
idsdB
r
4oiB
R
334
4 16
oo
i iB
R R
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Example Problem WS 11 #7 Two long, parallel wires separated by a distance of 6.0 cm carry current as shown. Both wires carry a current of 17.5 A. Point P is located 1/3 the distance from the left wire and
3.0 cm above the plane through both wires. What is the magnetic field (magnitude and direction)
at point P? We begin by using the equations we derived for
the long straight wires in order to calculate the
magnitudes of the magnetic fields. We have already determined the directions of the
magnetic fields at point P.
1R 2R
x
P
1B
2B
112
oiBR
222
oiBR
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Example Problem WS 11 #7
1R 2R
x
P
1B
2B
112
oiBR
222
oiBR
y
1sin y
R
1 sinyR
1
sin
2oiB
y
2sin y
R
2 sinyR
2
sin
2oiB
y
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Example Problem WS 11 #7
1
sin
2oiB
y
2
sin
2oiB
y
Vector Magnitude Angle x component y component Quad
-------------- -------------- -------------- -------------- -------------- --------------
1tan/ 3
y
x
1tan
2 / 3
y
x
P
1B
2B
RB
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Two wires carry currents as shown below. Calculate the force of attraction or repulsion that exists between these two wires. First let us determine the direction of the magnetic field around these two wires by using our right hand rule. Recall that the equation used to find the force on a charge carrying conductor is as follows. Lets consider the force on wire 2 due to wire 1. The magnetic field generated by wire 1 causes the force experienced by wire 2. Note the direction of this magnetic field acting on wire 2. Applying the cross product, we observe that
the force acting on wire 1 is directed
towards wire 2. Likewise, the force on wire 2 is directed
towards wire 1.
Force Between two Wires WS 11 #8 & 9
F iL B
1i
2i
1B
2B
2F
2F
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What do you suppose would happen if the current in one of the wires changes direction? Applying the cross product below and the right hand rule, we find that the force generated by
the magnetic fields cause the wires to move away from each other. In summary, if the current in both wires moves in the same direction, then the wires will attract
each other. If the current in the wires moves opposite direction, then the wires will repel each other.
2B
Force Between two Wires WS 11 #10 (Homework)
F iL B
1i
2i
1B
2F
1F
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P
I
WS 11 #12 (Homework) The figure to the right has an outer radius
r1 = 16.50 cm and an inner radius r2 = 8.85 cm.
The arc subtends an angle of 98.0. The current flowing through the object is
I = 0.523 A. Determine the magnetic field (magnitude
and direction) acting at point P.
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WS 11 #13 The figure below shows different arrangements of four cross-sectional views of
four current carrying wires. Which, if any, of these has a net zero magnetic field at the center? Explain your reasoning.
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WS 11 #14 (Homework) The figure to the right shows a current carrying conductor. This conductor splits into a circular and square segment shown. The current I = 3.80 A before it splits. The circle has a radius of 2.54 cm and the square has sides equal to 2.54 cm. What is the magnetic field (magnitude and direction) at the center of these two
shapes?
I2
I
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Ampere’s law is used to determine the net magnetic field produced by any distribution of current carrying conductors.
Consider the three cross sections of current carrying wires below. We use an Amperian Loop in order to determine the net magnetic field generated
by the enclosed wires.
Ampere’s Law in equation form is as follows where ienc is the net current enclosed by the Amperian loop.
The direction of the arrows on the Amperian loop are randomly chosen. Likewise, the vector B and angle are randomly chosen. In order to determine the magnitude of the currents in the wires enclosed b the
Amperian loop, we will use another right hand rule. Place your hand in a position where your fingers point in the same
direction as the arrows on the loop. If the current in the wire points in the same direction as
your thumb, then the current in that wire is
considered to be positive.
Ampere’s Law WS 11 #15 & 16
0 encB ds i
1i2i
3i
ds
B
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As a result, i1 would be positive and i3 would be negative.
Just as we did with Gauss’ law, we use symmetrical bodies as our Amperian surfaces.
Ampere’s Law WS 11 #15 & 16
0 encB ds i
1i2i
3i
ds
B
1 2enci i i
0 1 2B ds i i
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Find the magnetic field produced by the current carrying wire of radius r shown below.
What is the angle between ds and B? Do not forget the right hand rule for the Amperian surface.
Ampere’s Law – Long Straight Wire WS 11 #17
0 encB ds i
i
B
ds
r
2
00cos
r
encB ds i
2
00
r
encB ds i
2
00
r
encB s i
02 encrB i
0
2enci
Br
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Whereas an electric field cannot exist inside a current carrying conductor, a magnetic field can.
Find the magnetic field a distance of r from the center of a current carrying conductor of radius R (r < R).
Only a fraction of the total current passes through our Amperian surface
Ampere’s Law – Inside a Long Straight Wire WS 11 #18
0 encB ds i
i
B
ds
r
2
00cos
r
encB ds i
R2
2enc
ri
R
0
2enci
Br
20
22
rB
r R
022
rB
R
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Other Magnetic Fields We have already seen the magnetic field generated by a current carrying wire. In the next two slides we will see the magnetic fields generated by different
geometries of current carrying conductors.
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Other Magnetic Fields A coiled wire is
known as a solenoid. The magnetic field
generated by a solenoid looks like the one appearing below.
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Other Magnetic Fields
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P
-
+
r
P
R
2
1
3
P
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Other Magnetic Fields a
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Other Magnetic Fields a
P
I
I2
I
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1i2i
3i
ir
ir
R
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1i2i
1i2i