Download B . ds

Fig 30-CO, p.927
Ch 30 Sources of the Magnetic Field
30.1 The Biot-Savart Law



dB = 0 Ids x r /4r2
(cross or vector product)



dB = 0 Ids x r /4r2
Fig 30-1, p.927
Long Thin Wire (a << L)
a
r

I
x=0

r

ds = dx
x
x axis
B = 0I/2a
CT1: The magnitude of the magnetic fields are the
same at the outer loop in each diagram below and
the drawing is to scale. The ratio I1/I2 is
I1
I2
A. four
B. two C. one D. one-half E. one-quarter
P30.5 (p.859)
Ch 30 Sources of the Magnetic Field
30.2 The Magnetic Force between
Two Parallel Current-Carrying
Conductors
F/l = 0I1I2/2a
opposite direction currents repel
same direction currents attract
FT
P30.17 (p.860)
FL
X
B1
nonuniform
FB
FR
Ch 30 Sources of the Magnetic Field
30.3 Ampère’s Law
 B·ds = 0Inet
Arbitrary Current Carrying Loop (yellow)
ds
B
rd
d
r
I
B·ds = Brd
Ch 30 Sources of the Magnetic Field
30.3 Ampère’s Law
Method for Ampère’s Law: Use symmetry
Choose an Amperian loop such that on
a portion of the loop :
i. B || ds and B is constant.
ii. B.ds is zero because B = 0.
iii. B.ds is zero because B  ds.
iv. Any combination of i-iii.
P30.31 (p.862)
Amperian Loops
Coaxial Cable End View
I out of plane
I into plane
CT2: Use Ampere’s Law to determine the
magnetic field.
The field between the wire and shielding is
A. 0I/2r clockwise
B. 0I/2r counterclockwise
C. 0I/2r clockwise
D. 0Ir counterclockwise
E. zero
Coaxial Cable End View
I out of plane
I into plane
CT3: Use Ampere’s Law to determine the
magnetic field.
The field outside the shielding is
A. 0I/2r clockwise
B. 0I/2r counterclockwise
C. 0I/2r clockwise
D. 0Ir counterclockwise
E. zero
Ch 30 Sources of the Magnetic Field
30.4 The Magnetic Field of a Solenoid
B for an ideal solenoid: B = 0 outside and uniform inside.
B = 0nI
n = turns/length
Toroid: B = 0IN/2r
N = number of turns
P30.37 (p.863)
Ch 30 Sources of the Magnetic Field
30.5 Magnetic Flux and Gauss’s Law
in Magnetism
B =  B·dA = 0
Units: Tm2 = Weber (Wb)
Recall
E =  E·dA = qin/0
P30.39 (p.863)
P30.42 (p.863)
CT4: On a computer chip, two conducting strips
carry charge from P to Q and from R to S. If the
current direction is reversed in both wires, the
net magnetic force of strip 1 on strip 2
A. remains the
same.
B. reverses.
C. changes in
magnitude, but
not in direction.
D. changes to some
other direction.
E. other

Wrong Answer/Right Reason: Magnets attract other
magnets and some metals.
Right Answer/Wrong Reason: Magnets only interact with
other magnets.
Wrong Answer/Wrong Reason: Magnets are objects with
permanent electric charge on their ends.
CT5: Cosmic rays (atomic nuclei stripped bare of
their electrons) would continuously bombard Earth’s
surface if most of them were not deflected by Earth’s
magnetic field. Given that Earth is, to an excellent
approximation, a magnetic dipole, the intensity of
cosmic rays bombarding its surface is greatest at the
A. poles.
B. mid-latitudes.
C. equator.
CT6: A rectangular loop is placed in a
uniform magnetic field with the plane of
the loop perpendicular to the direction of
the field. If a current is made to flow
through the loop in the sense shown by the
arrows, the field exerts on the loop:
A. a net force.
B. a net torque.
C. a net force and a net torque.
D. neither a net force nor a net torque.
CT7: A rectangular loop is placed in a
uniform magnetic field with the plane of
the loop parallel to the direction of the
field. If a current is made to flow through
the loop in the sense shown by the arrows,
the field exerts on the loop:
A. a net force.
B. a net torque.
C. a net force and a net torque.
D. neither a net force nor a net torque.