SP212 Lesson 24 - United States Naval Academy · Je rey Larsen SP212 Lesson 24 March 6, 2018 3 / 8....

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SP212 Lesson 24 Ch. 29, Section 4 – B fields in solenoids and coils Jeffrey Larsen March 6, 2018 Jeffrey Larsen SP212 Lesson 24 March 6, 2018 1/8

Transcript of SP212 Lesson 24 - United States Naval Academy · Je rey Larsen SP212 Lesson 24 March 6, 2018 3 / 8....

Page 1: SP212 Lesson 24 - United States Naval Academy · Je rey Larsen SP212 Lesson 24 March 6, 2018 3 / 8. The Solenoid One of the more commonly encountered magnetic systems is the solenoid

SP212 Lesson 24Ch. 29, Section 4 – B fields in solenoids and coils

Jeffrey Larsen

March 6, 2018

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Page 2: SP212 Lesson 24 - United States Naval Academy · Je rey Larsen SP212 Lesson 24 March 6, 2018 3 / 8. The Solenoid One of the more commonly encountered magnetic systems is the solenoid

Aside: Finding NET B fields.

Sometimes we want to find a B field at point P in space with SEVERALcontributions to the field. It is just an exercise in adding vectors.

1 Each wire has a B field of Bi = µ0ii2πri

2 Each Bi field has a direction at P (CW/CCW) by RHR2.

3 At the same time, each Bi field is perpendicular to ri in space.

EXAMPLES:

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Challenge: More Ampere’s Law!

This is testing the direction of B vs. the direction of the loop. In the figure we seefour identical currents i and five Amperian loops (a through e) encircling them.

Rank the paths according to the value of∮~B · d~s taken in the directions shown,

most positive first.

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The Solenoid

One of the more commonly encountered magnetic systems is the solenoid - a long,tightly wound helical coil of wire.

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The Solenoid

The solenoid is useful because it produces a very uniform field inside.

Use Ampere’s law with a rectangular loop enclosing PART of the coil.

∮~B · d~s = µ0ienc

But our rectangular loop has four parts!∮~B · d~s =

∫ab~B · d~s +

∫bc~B · d~s +

∫cd~B · d~s +

∫da~B · d~s

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The Solenoid

Evaluate the four parts one at a time:∫ab~B · d~s = Bh∫

bc~B · d~s = −

∫da~B · d~s (they cancel)∫

cd~B · d~s = 0 (No B field outside)

So put that together....∮~B · d~s = Bh = µ0iencl

My loop encloses N turns of wire, each carrying a current in the same direction.Bh = µ0Ni

B = µ0Nh i = µ0ni

Where n is the number of turns per unit length of the solenoid

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The Magnetic Field of a Toroid

If we bend a solenoid around until it forms a closed ring, we have a toroid. h isthe circumference of the ring. The magnetic field inside a toroid is

B =µ0iN

2πr

Note that the field of a toroid is not uniform through its cross section, and N isnow the total number of turns, not the turns per unit length.

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The Field of a Current-Carrying Loop

Earlier, we showed that a current carrying loop had a dipole moment and actedlike a bar magnet, aligning with an external magnetic field. Also like a bar magnet,we now see that the current must create its own magnetic field. That field is:

~B(z) =µ0~µ

2πz3

Where z is the distance from the center of the loop, measured along the centralaxis. Don’t confuse ~µ = Ni~A (the dipole moment), with µ0, the permeability offree space!

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