OBJECTIVE:  I will be able to calculate the forces acting on the charges in a non-uniform wire using integration after today's class.

WORDS/FORMULAE FOR TODAY

TERMS

• Magnetic Field (B): A vector value
• Magnetic Force (F_B): Also a vector

CONSTANTS:

 

UNITS:

• Tesla = T defined as 1 N/C(m/s)

FORMULAE:

• F_B = qv x B
• F_B = qvB
• F_B = qvBsinθ
• F_B = IL x B
• F_B = ∫I ds x B NOTE: AP Version is: F_B = ∫I dℓ x B
• v = E/B
• KE = q₂B₂R₂/2m

WORK O' THE DAY: 

There are some kinda ugly derrivations in section 29.4 there but it essentially boils down to these ideas:

Imagine a wire where electrons are freely moving in an electrical current passing along the surface of that wire.

If that wire is oriented so that the wire is perpindicular to a magnetic field, it is reasonable to say that each and every electron in that wire experiences a force on the wire given by

F_B = qv x B

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Consider a usual household current of 10.0 amps... by definition that means there are 10.0 coloumbs of charge passing a fixed point every second.

Since 1.00 C of charge is defined to be 1.6 x 10¹⁹ electrons that means that every second there are

10.0 x 1.6 x 10¹⁹ electrons

passing through that wire each experiencing qv x B amount of force (which they in turn exert on the wire).

If you think this is sorta kinda like an integral is coming soon, YOU'RE RIGHT!

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Counting a huge amount of electrons to use in a calculation is obviously impractical...so we turn to integration.

Take a look at eq 29.10:

F_B = IL x B

magnetic force = (current)(wire length) x (magnetic field)

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Let's try a *SHORT* diff eq by evaluating the magentic force at an infintesimily small part of the wire:

ds = infintessimal segment of the wire

dF_B = the magnetic force exerted on an electron at that segment

dF_B = I ds x B

Time to integrate:

∫dF_B = I ∫ds x B (since current is constant)

which gives us:

F_B = I ∫ds x B

Notice the importance of using the X product since we are including the vector quantity and NOT just the magnitude for magentic force (F_B)

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KEY BITS! Does the problem ask for "magnitude" ONLY?

If yes then

GOOD NEWS, NO CROSS-PRODUCTING

Does the problem ask the "magnetic force"? (sometimes we can luck out when the problem includes unit vectors-- that's usually a clue to go cross-producting but that's not a for sure kinda thing)

BAD NEWS-- GOTTA CROSS PRODUCT

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Please work through example 29.4... Sketch the problem, list initial conditions and formulae and then try it (Hint: the problem asks for Magnitude AND Direction which means.... WHAT?). Annotate with any questions that you have during your first run through.

Refer back to the book when (and if) you get stuck

Note the solution and add that to your work with appropriate annotation(s).

Work until you get stuck again, then refer to the book, find the next step, add that to your solution with appropriate annotations.

COURSEWORK:29.32, 29.35, 29.39

ANSWERS: