© 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


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


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


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


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


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


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


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


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


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


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


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


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.


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.


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


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


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


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


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


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.


Other Magnetic Fields A coiled wire is

known as a solenoid. The magnetic field

generated by a solenoid looks like the one appearing below.


Other Magnetic Fields


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P

-

+

r

P

R

2

1

3

P


Other Magnetic Fields a


Other Magnetic Fields a

P

I

I2

I


1i2i

3i

ir

ir

R


1i2i

1i2i

© 2001-2005 shannon w. helzer. all rights reserved. 1 chapter 29 – magnetic fields due to current

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