Announcements • WebAssign HW Set 5 due October 10 • Problems cover material from Chapters 18 • HW set 6 due on October 17 (Chapter 19) • Prof. Kumar tea and cookies Tuesdays from 5 – 6… pm in room 2165 • Exam 1 statistics average 12.78 stand. dev. 3.51
Announcements WebAssign HW Set 5 due October 10 Problems cover material from Chapters 18
Jan 02, 2016
Announcements WebAssign HW Set 5 due October 10 Problems cover material from Chapters 18 HW set 6 due on October 17 (Chapter 19) Prof . Kumar tea and cookies Tuesdays from 5 – 6… pm in room 2165 Exam 1 statistics average 12.78 stand. dev. 3.51 QUESTIONS? PLEASE ASK!. - PowerPoint PPT Presentation
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Announcements
• WebAssign HW Set 5 due October 10• Problems cover material from Chapters 18
• HW set 6 due on October 17 (Chapter 19)
• Prof. Kumar tea and cookies Tuesdays from 5 – 6… pm in room 2165
• Exam 1 statistics average 12.78
stand. dev. 3.51
QUESTIONS? PLEASE ASK!
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From last time Magnets and earth’s magnetic
field
Magnetic Fields:
Units are T = N/A.m Use right hand rule to determine
direction of force
Force on a wire: F = B I L sin θ
Continuing Forces and torques
on current carrying wires and loops.
On moving charges. Magnetic Fields and
Amperes law. Forces between two
current carrying wires. d
II
l
F
IlB
r
IB
qBmvr
BIANB
o
o
o
2
.
2
/
sinsin
21
Torque on a Current Loop Torque t = B I A N sin q
Applies to any shape loop N is the number of turns in the
coil Torque has a maximum value
of NBIA (when q = 90°) Torque is zero when the field is
parallel to the plane of the loop
Magnetic Moment m = IAN m is a vector Torque can be written as
t = mB sinq
m
Example Problem 19.31
A long piece of wire with a mass of 0.100 kg and a length of 4.00 m is used to make a square coil with a side of 0.100 m. The coil is hinged along a horizontal side, carrying a 3.40 A current, and is placed in a vertical magnetic field of 0.010 T. (a) Determine the angle that plane of the coil makes with the vertical when the coil is in equilibrium. (b) Find the torque acting on the coil due to the magnetic force at equilibrium
Electric Motor
electric motor - converts electrical energy to mechanical energy The mechanical energy is
in the form of rotational kinetic energy
An electric motor consists of a rigid current-carrying loop that rotates when placed in a magnetic field
Electric Motor
Torque acting on the loop will rotate the loop to smaller values of θ until the torque becomes 0 at θ = 0°
If the loop turns past this point and the current remains in the same direction, the torque reverses and turns the loop in the opposite direction Bad!!
Electric Motor
So, we need to be clever…
To provide continuous rotation in one direction, the current in the loop must periodically reverse In AC motors, this reversal
naturally occurs In DC motors, a split-ring
commutator and brushes are used
Actual motors would contain many current loops and commutators
Force on a Charged Particle in a Magnetic Field
Consider a particle moving in an external magnetic field so that its velocity is perpendicular to the field
The force is always directed toward the center of the circular path
The magnetic force causes a centripetal acceleration, changing the direction of the velocity of the particle
Force on a Charged Particle
Equating the magnetic and centripetal forces:
Solving for r:
r is proportional to the momentum of the particle and inversely proportional to the magnetic field
Sometimes called the cyclotron equation
Particle Moving in an External Magnetic Field
If the particle’s velocity is not perpendicular to the field, the path followed by the particle is a spiral The spiral path is
called a helix
Example Problem 19.42
A cosmic ray proton in interstellar space has an energy of 10 MeV and executes a circular orbit having a radius equal to that of Mercury’s orbit around the Sun (5.8 x 1010 m). What is the magnetic field in that region of space?
Solution to 19.31
Solution to 19.42
Example Problem 19.42
A cosmic ray proton in interstellar space has an energy of 10 MeV and executes a circular orbit having a radius equal to that of Mercury’s orbit around the Sun (5.8 x 1010 m). What is the magnetic field in that region of space?
Magnetic Fields – Long Straight Wire
A current-carrying wire produces a magnetic field B Right hand rule # 2 to
determine direction of B
Magnitude of B at a distance r from a wire carrying current of I is:
µo = 4 x 10-7 T.m / A µo is called the permeability
of free space
Ampère’s Law: General relationship between I in
an arbitrarily shaped wire and B produced by the wire:
B|| Δℓ = µo I
Choose an arbitrary closed path around the current
Sum all the products of B|| Δℓ around the closed path
Ampère’s Law Applied to a Long Straight Wire
Use a closed circular path
The circumference of the circle is 2 r
This is identical to the result previously shown
Example Problem 19.54
Two long parallel wires separated by a distance 2d carry equal currents in the same direction. The currents are out of the page in the figure. (a) What is the direction of the magnetic field at P on the x-axis set up by the two wires? (b) Find an expression for the magnitude of the field at P. (c) From (b), determine the field midway between the two wires.
Magnetic Force Between Two Parallel Conductors
The force on wire 1 is due to the current in wire 1 and the magnetic field produced by wire 2
The force per unit length is:
Parallel conductors carrying currents in the same direction attract each other
Parallel conductors carrying currents in the opposite directions repel each other
Magnetic Field of a Current Loop
The magnitude of the magnetic field at the center of a circular loop with a radius R and carrying current I is
With N loops in the coil, this becomes:
Magnetic Field of a Solenoid Solenoid – long straight wire is
bent into a coil of several closely spaced loops
Electromagnet - acts like a magnet only when it carries a current
B field lines inside the solenoid are nearly parallel, uniformly spaced, and close together B is nearly uniform and strong
The exterior field is nonuniform, much weaker, and in the opposite direction to the field inside the solenoid
Magnetic Field in a Solenoid, Magnitude
The magnitude of the field inside a solenoid is constant at all points far from its ends
B = µo n I n is the number of turns
per unit length n = N / ℓ
The same result can be obtained by applying Ampère’s Law to the solenoid
Example Problem 19.60
A certain superconducting magnet in the form of a solenoid of length 0.5 m can generate a magnetic field of 9.0T in its core when the coils carry a current of 75 A. The windings, made of a niobium-titanium alloy, must be cooled to 4.2K. Find the number of turns in the solenoid.
Solution to 19.42
Solution to 19.54
Solution to 19.60
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