admin: mid-term 2 grades and solutions posted. scripts back at the end average = 84% (up from 72%)....
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Admin:• mid-term 2 grades and solutions posted. Scripts back at the end• Average = 84% (up from 72%). Excellent work!• Assignment 10 posted.
• Due on Monday (May 18th) • Missed assignments can be completed up until the final
(May 22nd).• The class is over after the final exam
• no extensions or late assignments after this.• Any issues with grades (labs, assignments, mid-terms) must
be resolved before then with me or your TA.
F 0D- 59.5D 62.5D+ 66.5C- 69.5C 72.5C+ 76.5B- 79.5B 82.5B+ 86.5A- 89.5A 92.5
72% 84% 75% 89% 93%10.82 12.56 22.50 17.70 18.55 82.13
Mt1 (15%) Mt2 (15%) Final (30%) MastPhys (20%) Labs (20%)AveragePoints
Grading scheme(and how to calculate your grade)
Copyright © 2009 Pearson Education, Inc.
Chapter 27Magnetism
Copyright © 2009 Pearson Education, Inc.
A uniform magnetic field is constant in magnitude and direction.
The field between these two wide poles is nearly uniform.
27-1 Magnets and Magnetic Fields
Copyright © 2009 Pearson Education, Inc.
Experiment shows that an electric current produces a magnetic field. The direction of the field is given by a right-hand rule.
27-2 Electric Currents Produce Magnetic Fields
Copyright © 2009 Pearson Education, Inc.
27-2 Electric Currents Produce Magnetic Fields
Here we see the field due to a current loop; the direction is again given by a right-hand rule.
Copyright © 2009 Pearson Education, Inc.
A magnet exerts a force on a current-carrying wire. The direction of the force is given by a right-hand rule.
27-3 Force on an Electric Current in a Magnetic Field; Definition of B
Copyright © 2009 Pearson Education, Inc.
Internet hints…
Copyright © 2009 Pearson Education, Inc.
The force on the wire depends on the current, the length of the wire, the magnetic field, and its orientation:
This equation defines the magnetic field B.
In vector notation :
27-3 Force on an Electric Current in a Magnetic Field; Definition of B
Copyright © 2009 Pearson Education, Inc.
Vector Cross Product;
The vector cross product is defined as:
The direction of the cross product is defined by a right-hand rule:
http://www.mrbigler.com/moodle/mod/resource/view.php?id=3721
Copyright © 2009 Pearson Education, Inc.
11-2 Vector Cross Product; Torque as a Vector
Some properties of the cross product:
Copyright © 2009 Pearson Education, Inc.
Unit of B: the tesla, T:
1 T = 1 N/A·m.
Another unit sometimes used: the gauss (G):
1 G = 10-4 T.
27-3 Force on an Electric Current in a Magnetic Field; Definition of B
Copyright © 2009 Pearson Education, Inc.
Just for interest…
Copyright © 2009 Pearson Education, Inc.
27-3 Force on an Electric Current in a Magnetic Field; Definition of B
Example 27-1: Magnetic Force on a current-carrying wire.
A wire carrying a 30-A current has a length l = 12 cm between the pole faces of a magnet at an angle θ = 60°, as shown. The magnetic field is uniform at
0.90 T. What is the magnitude of the force on the wire?
Copyright © 2009 Pearson Education, Inc.
27-3 Force on an Electric Current in a Magnetic Field; Definition of B
Example 27-2: Measuring a magnetic field.
A rectangular loop of wire hangs vertically as shown. A magnetic field B is directed horizontally, perpendicular to the wire, and points out of the page at all points. The magnetic field is very nearly uniform along the horizontal portion of wire ab (length l = 10.0 cm) which is near the center of the gap of a large magnet producing the field. The top portion of the wire loop is free of the field. The loop hangs from a balance which measures a downward magnetic force (in addition to the gravitational force) of F = 3.48 x 10-2 N when the wire carries a current I = 0.245 A. What is the magnitude of the magnetic field B?
Copyright © 2009 Pearson Education, Inc.
Example 27-3: Magnetic Force on a semicircular wire.
A rigid wire, carrying a current I, consists of a semicircle of radius R and two straight portions as shown. The wire lies in a plane perpendicular to a uniform magnetic field B0. Note choice of x and y axis. The straight portions each have length l within the field. Determine the net force on the wire due to the magnetic field B0.
See the book for the full solution (there are some tricks)
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