Lesson 21-3 Magnetic Force

8/2/2019 Lesson 21-3 Magnetic Force

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Lesson 21-3 Magnetic Force

Homework

Read pages 773-780 in Holt text.

Complete lesson 7-3 in HomeworkHelpers: Physics.

AP Physics Read pages 644-654 in

College Physics. Check Moodle for additional assignments

from this section.


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Section Objectives

Given the force on a charge in a magneticfield, determine the strength of themagnetic field.

Use the right-hand rule to find the directionof the force on a charge moving through amagnetic field.

Determine the magnitude and direction ofthe force on a wire carrying current in amagnetic field.


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When a charged particle enters amagnetic field, it experiences a force.


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The force experienced by the particle isat a right angle to both its velocity and

the magnetic field.

Movie Clip
http://www.fordhamprep.org/moodle/file.php/32/PowerPoint_Lessons/Chapter_21_Magnetism/Lesson_21-3_Magnetic_Force/f=qvb%20and%20right%20hand%20rule.MOV


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This deflection of charged particles by magneticfields is the basis behind CRT (cathode-raytube) monitors and T.V.s.


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This effect can even beused to propel a

submarine through theocean.


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The direction of this force can bedetermined by the extended right-hand

rule.


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The direction ofthis force can be

determined bythe extendedright-hand rule.


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Find the direction of the magnetic field (B)in each of the following diagrams.


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The magnetic force willdeflect the charged particle.

Because the charged particlechanges direction, thedirection of the force on italso changes. This results in

circular motion.


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Click on the image below to see a movingcharge enter a magnetic field.

Movie Clip
http://www.fordhamprep.org/moodle/file.php/32/PowerPoint_Lessons/Chapter_21_Magnetism/Lesson_21-3_Magnetic_Force/charged%20particle%20in%20magnetic%20field.MOV


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Perhaps, you can see how Earths magneticfield directs charged particles towards thepoles, producing auroras.

Movie Clip
http://www.fordhamprep.org/moodle/file.php/32/PowerPoint_Lessons/Chapter_21_Magnetism/Lesson_21-3_Magnetic_Force/charged%20particle%20in%20earth's%20magnetic%20field.MOV


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The extended right-hand rule can also beused to find the direction of the forceexperienced by a current-carrying wire in amagnetic field.


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Click on the image below to see anexample of the force experienced by acurrent-carrying wire in a magnetic field.

Movie Clip
http://www.fordhamprep.org/moodle/file.php/32/PowerPoint_Lessons/Chapter_21_Magnetism/Lesson_21-3_Magnetic_Force/force%20on%20current%20carrying%20wire.MOV


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Click on the image below to try Physletillustration 27.4: Magnetic Forces on

Currents.
http://fordhamprep.org/physics/physlets/contents/electromagnetism/b_fields/illustration27_4.html


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Imagine a current-carrying wire loop ina magnetic field.


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The hand-rule shows an opposite forceon each side.


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These opposite forces produce torque.


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This torque is the basisfor common devices,such as thegalvanometer, motorand generator.


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The magnitude of the force that a chargedparticle experiences when moving through amagnetic field is given by the formula below.

BSinqFB

Where; q = charge, = velocity of the charge,B = magnetic field strength and is =the anglebetween the field and the velocity vector.


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When the velocity of the particle isperpendicular to the field, the simplifiedversion of the formula can be used, becausesin 90o = 1

BqFB


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The angle between the velocity vector andthe magnetic field must be taken intoaccount.


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Lesson 21-3 Magnetic ForceExample 1. A proton entering a magnetic field with avelocity of 3.7 m/s at an angle of 55.0o experiences aforce of 3.4 x 10-3 N. Find the strength of the field.


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Lesson 21-3 Magnetic ForceExample 1. A proton entering a magnetic field with avelocity of 3.7 m/s at an angle of 55.0o experiences aforce of 3.4 x 10-3 N. Find the strength of the field.

Given:

Find:

Solution:

q = 1.60 x 10-19 C = 3.7 m/s = 55.0o FB = 3.4 x 10

-3 N

B

sinq

F

BB

))5.05m/s)(sinC)(3.710x((1.60

N10x3.4

o19-

-3

= 7.0 x 1015 T


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The magnitude of the force that a current-carrying wire experiences in a magnetic fieldis given by the formula below.

Where; B = magnetic field strength, I = current,= length, and is =the angle between the field

and the orientation of the wire.

SinBIFB


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When the wire is perpendicular to the field,the simplified version of the formula may beused.

BIFB


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Example 2. A wire segment with a length of 1.3 mhas a steady current of 6.2 A through it. What is themagnitude of the force it would experience in a

magnetic field measuring 2.3 x 10-4

T if the wire isorientated 7.5o to the magnetic field?


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Example 2. A wire segment with a length of 1.3 m hasa steady current of 6.2 A through it. What is themagnitude of the force it would experience in a magnetic

field measuring 2.3 x 10-4

T if the wire is orientated 7.5o

to the magnetic field?

Given:

Find:

Solution:

I = 6.2 A = 1.3 m B = 2.3 x 10-4 T = 7.5o

FB

sinBIFB

= (2.3 x 10-4 T)(6.2 A)(1.3 m)(sin 7.5o)

= 2.4 x 10-4 N


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Two parallel current-carrying wires exert onone another forces thatare equal in magnitudeand opposite in direction.


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If the currents are in the same direction, thewires will attract.


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If the currents are in the opposite direction,the wires will repel.


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Summary

The extended right-hand rule can be used

to determine the direction of the forceexperienced by a charged particle movingthrough a magnetic field.

The extended right-hand rule can also be

used to determine the direction of the forceexperienced by a current-carrying wire in amagnetic field.


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Summary - Formulas

BSinqFB SinBIFB

Documents

Lesson 21-3 Magnetic Force