June 2011 G485-Fields, Particles and Frontiers of Physics Watch

anshul95
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(Original post by Lengalicious)
- Protons posses magnetic property as they spin on their axis.
- Application of magnetic field causes proton to precess about its axis due to the magnetic interaction.
- The frequency of precession = the larmor frequency.
- If the frequency of AC in the radio coil = the precessional frequency the proton will resonate.
- Nuclear resonance causes the proton to gain energy boosting it up to a higher unstable energy level.
- The proton will lose this energy as radio waves after a certain time known as nuclear relaxation.
- Radio wave picked up by radio coils and interpreted by computer to form 3D image on screen.
- A series of smaller magnets are placed in the main magnet of the scanner to alter the otherwise uniform magnetic field so the relaxation times of the emitted Rwaves are different so the data can be interpreted.
the last line is incorrect. The relaxation times only depend on the hydrogen environments i.e. tissue types. The gradient coil changes the lamour frequency of the protons so that the exact position of resonance can be determined by measuring the frequency of the returning RF radiation.
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Ralphus J
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Will we have to know how to calculate larmor frequency?
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sulexk
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(Original post by Ralphus J)
Will we have to know how to calculate larmor frequency?
Larmor frequency and precession frequency are related, however the larmor frequency is the angular frequency of precession W=yB

where y=gyromagnetic ratio - depends on nucleus
B= magnetic flux density- of external magnetic field

Precession frequency = Larmor frequency / 2pi

I know, it seems complicated.

But to not complicate things, if they ask for precession frequency use that formula, if they ask for larmor frequency just workout W=yB

Think about it like this - how do you workout frequency x/2pi, where x is number of radians covered per second. when you divide this by 2pi you get the number of circles traversed per second. So you can see that the larmour frequency is the "angular frequency of precession".

Hope that helps- please donot hesitate to reply.

Thank you
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Ralphus J
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Thanks SulexK !

What does anshul 95 mean by "AC" what does that stand for? alternating current?
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sulexk
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(Original post by Ralphus J)
Thanks SulexK !

What does anshul 95 mean by "AC" what does that stand for? alternating current?
Hmm, I think he means that if the frequency of the radio pulse sent out by the RF transmitter coil into the patient is equal to the precessional frequency then the protons shall resonate

Hope that helps!
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Ralphus J
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Sorry to keep hassling on but im still not entirely sure on these two, are they correct, have i missed out anything? Thanks alot


• Nuclear magnetic resonance – Protons precessing about the strong external field are exposed to a burst or pulse of RF waves whose frequency equals the frequency of precession. Each proton absorbs a photon of RF energy and flips into a higher energy state.

• Nuclear relaxation – When the external radio waves stop and the protons gradually relax into their lower energy states, as they do they release their excess energy as RF waves. The relaxation times only depend on the hydrogen environments i.e. tissue types.
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Oh my Ms. Coffey
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(Original post by Ralphus J)
Sorry to keep hassling on but im still not entirely sure on these two, are they correct, have i missed out anything? Thanks alot


• Nuclear magnetic resonance – Protons precessing about the strong external field are exposed to a burst or pulse of RF waves whose frequency equals the frequency of precession. Each proton absorbs a photon of RF energy and flips into a higher energy state.

• Nuclear relaxation – When the external radio waves stop and the protons gradually relax into their lower energy states, as they do they release their excess energy as RF waves. The relaxation times only depend on the hydrogen environments i.e. tissue types.

Protons precess at Larmour frequency around the magnetic field lines but not in phase, when a radio frequency coil is used at the Larmour phase also they precess in phase and also flip over, when the radio frequency is removed then they start to relax emitting electromagnetic radiation.
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anshul95
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(Original post by Ralphus J)
Thanks SulexK !

What does anshul 95 mean by "AC" what does that stand for? alternating current?
I didn't actually write that - I only corrected the person who did.
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ViralRiver
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Can someone help me please?

On page 113 look at question 1, is the book's answer wrong?

My understanding is that if the coil is parallel to the field - no e.m.f is induced - not cutting field. However, when it is perpendicular to field it cuts magnetic field at maximum rate. If you look at the first paragraph (under stretch and challenge) on the same page it says the opposite.

Which is correct?
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sulexk
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(Original post by ViralRiver)
Can someone help me please?

On page 113 look at question 1, is the book's answer wrong?

My understanding is that if the coil is parallel to the field - no e.m.f is induced - not cutting field. However, when it is perpendicular to field it cuts magnetic field at maximum rate. If you look at the first paragraph (under stretch and challenge) on the same page it says the opposite.

Which is correct?
actually when the coil is parallel to the field, the rate of change of magnetic flux linkage will be maximum(think about the subsequent motion), and hence the induced EMF will be maximum. The when the coil is perpendicular to the field rate of change of flux linkage is zero( it is maximum flux linkage- hence rate of change of flux linkage is zero) so no induced EMF
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ViralRiver
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(Original post by sulexk)
actually when the coil is parallel to the field, the rate of change of magnetic flux linkage will be maximum(think about the subsequent motion), and hence the induced EMF will be maximum. The when the coil is perpendicular to the field rate of change of flux linkage is zero( it is maximum flux linkage- hence rate of change of flux linkage is zero) so no induced EMF
Ok so do you mean that when it is perpendicular to the field, it has maximum flux linkage but minimum rate of change of flux as gradient = 0?
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sulexk
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(Original post by ViralRiver)
Ok so do you mean that when it is perpendicular to the field, it has maximum flux linkage but minimum rate of change of flux as gradient = 0?
Absolutely!
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ChoYunEL
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(Original post by ViralRiver)
Can someone help me please?

On page 113 look at question 1, is the book's answer wrong?

My understanding is that if the coil is parallel to the field - no e.m.f is induced - not cutting field. However, when it is perpendicular to field it cuts magnetic field at maximum rate. If you look at the first paragraph (under stretch and challenge) on the same page it says the opposite.

Which is correct?
At 0 - it's cutting the field fully, +4v
Each second = 1 revolution
At 1 - +4v
Therefore at 0.5, the voltage is -4v

0.25 seconds is halfway through 0 and 0.5 therefore it's changed by pi/2
Voltage becomes 0,

Lets evaluate, at the following times...
0: +4v
0.25: 0v
0.5: -4v
0.75: 0v
1: +4v

A full period.

At
0: Parallel
0.25: pi/2 out , etc.

Question for those without the book
A small generator coil is rotated at 60 revolutions per minute in the uniform magnetic field between the poles of an electromagnet. The coil is connected via slip-rings to a resistor and oscilloscope.
At the instant shown, t=0, the oscilloscope reading is +4v. State the reading at
t = (i) 0.25s (ii) 0.50s (iii) 1.75s (iv) 3s
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ViralRiver
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(Original post by ChoYunEL)
At 0 - it's cutting the field fully, +4v
Each second = 1 revolution
At 1 - +4v
Therefore at 0.5, the voltage is -4v

0.25 seconds is halfway through 0 and 0.5 therefore it's changed by pi/2
Voltage becomes 0,

Lets evaluate, at the following times...
0: +4v
0.25: 0v
0.5: -4v
0.75: 0v
1: +4v

A full period.

At
0: Parallel
0.25: pi/2 out , etc.

Question for those without the book
A small generator coil is rotated at 60 revolutions per minute in the uniform magnetic field between the poles of an electromagnet. The coil is connected via slip-rings to a resistor and oscilloscope.
At the instant shown, t=0, the oscilloscope reading is +4v. State the reading at
t = (i) 0.25s (ii) 0.50s (iii) 1.75s (iv) 3s
Ok can you please make something clear to me. At t = 0, the coil is parallel to the field, this means that the flux linkage at that time is 0Wb right? However, this means that the rate of change of flux linkage, i.e the e.m.f induced is at its maximum, 4V? Thanks .
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ViralRiver
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(Original post by sulexk)
Absolutely!
Thanks .
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susan23
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this exam is on tuesday morning yh? just double checking not 2mrw?
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ChoYunEL
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(Original post by ViralRiver)
Ok can you please make something clear to me. At t = 0, the coil is parallel to the field, this means that the flux linkage at that time is 0Wb right? However, this means that the rate of change of flux linkage, i.e the e.m.f induced is at its maximum, 4V? Thanks .
Taking data from the formula booklet

Flux Linkage = B * A * CosX
Now, when the thing is parallel, in terms of a circle, the angle is 0
Cos0 = 1
When it is at 90, pi/2 CosX = 0

In addition, induced e.m.f = -rate of change of magnetic flux linkage
If you plot a cos graph, the point of greatest change (the gradient) = max induced e.m.f.

For this particular question, at t=0, there is no flux linkage since flux linkage is when the field is being cut. When the wire is inside the field, no fields are being cut. It is when it is rotating outside the field and the field is being cut the most when the wire is horizontal.

This question is plotted on a sin graph in terms of Flux Linkage.
dy/dx of SinX = CosX which is why you create a Cos graph for 1(b)

A particular question good for observing this explanation is page 122 question 2.
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ChoYunEL
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(Original post by susan23)
this exam is on tuesday morning yh? just double checking not 2mrw?
You are safe :P
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magdaplaysbass
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(Original post by M_I)
In the specimen paper, question 8.d)i) why do the multiply by 100?
dividing 18 by 0.15 will give you 1% of the power - which is 120
so you need to multiply it by 100 to get 100% of the power (12,000)
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FristyKino
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(Original post by ChoYunEL)
You can go to other exam boards and try their question...
That's what I went down to in Jan... it didn't help
Lol'd at this so hard at 6am in the morning - probably not the best thing to do at this time lol
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