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Magnetic Fields, Deuterium and curved tracks

  • Thread starter TFM
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  • #1
TFM
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[SOLVED] Magnetic Fields, Deuterium and curved tracks

Homework Statement



A deuteron (the nucleus of an isotope of hydrogen) has a mass of [tex] m_D [/tex] and a charge of e. The deuteron travels in a circular path with a radius of r in a magnetic field with a magnitude of B.

Find the time required for it to make 1/2 of a revolution.

Homework Equations



Cyclotron frequency: [tex] \omega = \frac{v}{R} [/tex]

The Attempt at a Solution



IO have already calculated the velocity in the previous part to be:

[tex] \frac{reB}{m_D} [/tex]

Frequncy is 1/Period, so I get

[tex] period = \frac{1}{\frac{v}{R}} = \frac{R}{v} [/tex]

[tex] = \frac{r}{\frac{reB}{m_D}} [/tex]

Which I have rearranged to get:

[tex]\frac{m_D}{eB} [/tex]

Sincve this is for one whol revolution, I multiplied it by a half to get:

[tex]\frac{m_D}{2eB} [/tex]

Which Mastering Physics says is wrong, but also says:

Your answer is off by a multiplicative factor.

Any Ideas?

TFM
 

Answers and Replies

  • #2
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T = 2*pi/omega
 
  • #3
TFM
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So that would make:

[tex] period = \frac{2\pi}{\frac{v}{R}} = \frac{2\piR}{v} [/tex]

and this the fraction would be:

[tex]\frac{2\pi m_D}{2eB} [/tex]

Does this look more right?

TFM
 
  • #4
TFM
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Does the above look correct now?

TFM
 
  • #5
44
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Yes.
 
  • #6
TFM
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Thanks,

The next part of the question asks:

Through what potential difference would the deuteron have to be accelerated to acquire this speed?

But I am niot sure what formula to use.

Any suggestions,

TFM
 
  • #7
44
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Use equation of equivalence of potential and kinetic energy (potential energy turns to kinetic)
 
  • #8
TFM
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Kinetic Energy:

[tex] K.E. = \frac{1}{2}mv^2 [/tex]

Electric Potential:

[tex] U= q_0V [/tex]

Equate:

[tex] \frac{1}{2}mv^2 = q_0V [/tex]

[tex] V = \frac{mv^2}{2q_0} [/tex]

Is this correct?

TFM
 
  • #9
TFM
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Putting in my values I get

[tex] \frac{4 \Pi^2m_D^3}{\frac{4e^2B^2}{2e}} [/tex]

Does this look correct?

TFM
 
  • #10
44
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Nope, there is no place for pi here v=omega*r.

using omega=eB/m you should get it
 
Last edited:
  • #11
TFM
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I looked at the wrong part :redface: - i had already found the speed to be:

[tex] v = \frac{reB}{m_D} [/tex]

so putting in this v, I get:

[tex] V = \frac{m_D(\frac{r^2e^2B^2}{m_D^2})}{2e} [/tex]

is this better?

TFM
 
  • #12
44
0
This should be correct.
 
  • #13
TFM
1,026
0
Indeed it is correct.

Thanks, michalll :smile:

TFM
 

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