- StudyBlue
- Tennessee
- University of Tennessee - Knoxville
- Physics
- Physics 231
- Jones
- University Physics Chapter 27 part 2

Anonymous

Advertisement

Let?s review ... ?Magnets always have a north and a south pole. There are no ?free? north or south poles. ?The needle of a compass will deflect in the presence of a wire with an electric current flowing through it (Oersted?s experiment). ?A magnetic field exerts a force on moving charges in that field. ?This force is perpendicular to both the magnetic field and the velocity vectors. ?The direction of the force is found using the right hand rule. ?The force is given by: ! r F =q r v " r B Let?s review cont. ?For a constant force: ?F is paralel to v, uniform aceleration ?F is perpendicular to v, uniform circular motion ?F is neither paralel or perpendicular, helical motion ?EM ?Lorentz? force for charged particles in electric and magnetic fields, is the vector sum of the electric and magnetic forces: ?A velocity selector is a device with crossed electric and magnetic fields, which allows only those charged particles with a certain velocity to pass through. This led to the discovery of the electron. )(BvEqF rr v !+= R mv aF 2 = r r tv+= 0 ! v= E B Magnetic force on a proton ?Example 27.1 Y&F ?A beam of protons (q=1.6 x 10 -19 C) moves at 3.0 x 10 5 m/s through a uniform magnetic field with magnitude 2.0 T that is directed along the positive z- axis. The velocity of each proton lies in the xz-plane at an angle of 30 o to the +z-axis. Find the force on the proton. ?Use ?Direction of force RH rule ?Magnitude of force ! r F =q r v " r B ! F=qvBsin" =(1.6#10 $19 C)(3.0#10 5 m/s)(2.0T)(sin30°) =4.8#10 $14 N Force on a curent carying lop ?Calculate the force using: ! r F =Id r l " r B # Each side experiences a force of magnitude: By symmetry the total force on the loop is zero, the forces on opposite sides (or ends) are equal and opposite. The net force on a current loop in a uniform magnetic field is zero. ! F=IaB Torque on a curent carying lop ?The net force on the closed lop is zero, so there is no translational motion ?However, there may be a net torque, which leads to rotational motion ! r " = r r # r F (se chapter 10 Y&F) In the y-direction, the forces F and -F are colinear, so there is no net torque; however this is not so in the x-direction. The lever arms asociated with the torque are: ! b 2 sin" ! "=2F b 2 # $ % & ? ( sin)=IBAsin) Hence, the net torque is around the y-axis and has a magnitude Magnetic dipole moment ?Torque is in the positive y direction and is given by ?The product IA is defined as the magnetic dipole moment or magnetic moment, for which we use the symbol µ ?Hence the torque can be rewritten as: ! "=IBAsin# ! µ=IA ! "=µBsin# Magnetic dipole moment ?The magnetic dipole moment can be written as a vector ?The direction of µ is therefore the same as the direction of A. The direction of A is defined by the sense of the current (use the right hand rule). ! r µ =I r A ! r " = r µ # r B Potential energy for a magnetic dipole ?The magnetic field does work on the loop as it turns due to the torque it experiences ?The potential energy of the loop is given by: ?The potential energy is 0 when the magnetic moment and the magnetic field vector are perpendicular ?The potential energy is smallest when µ and B are parallel (note the negative sign) ?The potential energy is largest when µ and B are anti-parallel ! U=" r µ ? r B Potential energy for a magnetic dipole ?Where did the equation for potential energy come from? ?Motion along a straight line ?Rotation around an axis ?If we want to turn the lop against the field then we have to aply an oposite torque ?An infinitesimal amount of work done by this external torque is ?The work done to rotate the moment from ? 0 to ? 1 is: ?Change in potential energy is given by U=-W, so ! "=µBsin# ! "=#µBsin$ ! dW="µBsin#d# ! dW=Fdl ! dW="d# ! W="(µB)sin#d# # 0 # 1 $ =µB(cos# 1 "cos# 0 ) ! U="µBcos# U=" r µ ? r B Solenoids ?The equationholds for any dipole irrespective of its shape e.g. for a circular loop ?If we wind multiple loops from a conducting wire, we have made a solenoid ! r µ =I r A ?The dipole moment is then the sum of the dipole moments of the loops (or turns) where N is the number of turns ?Like permanent magnets, solenoids and all magnetic dipoles will ?try? to align along the magnetic field ! r µ =NI r A System Administrator chapter27_part2.ppt

Advertisement

Want to see the other 10 page(s) in University Physics Chapter 27 part 2?
JOIN TODAY FOR FREE!

"The semester I found StudyBlue, I went from a 2.8 to a 3.8, and graduated with honors!"

Jennifer Colorado School of Mines
StudyBlue is not sponsored or endorsed by any college, university, or instructor.

© 2014 StudyBlue Inc. All rights reserved.

© 2014 StudyBlue Inc. All rights reserved.