Magnetic Fields, Deuterium and curved tracks

[TFM]

[SOLVED] Magnetic Fields, Deuterium and curved tracks

Homework Statement

A deuteron (the nucleus of an isotope of hydrogen) has a mass of $$m_D$$ 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: $$\omega = \frac{v}{R}$$

The Attempt at a Solution

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

$$\frac{reB}{m_D}$$

Frequncy is 1/Period, so I get

$$period = \frac{1}{\frac{v}{R}} = \frac{R}{v}$$

$$= \frac{r}{\frac{reB}{m_D}}$$

Which I have rearranged to get:

$$\frac{m_D}{eB}$$

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

$$\frac{m_D}{2eB}$$

Which Mastering Physics says is wrong, but also says:

Your answer is off by a multiplicative factor.

Any Ideas?

TFM

 

[michalll]

T = 2*pi/omega

 

[TFM]

So that would make:

$$period = \frac{2\pi}{\frac{v}{R}} = \frac{2\piR}{v}$$

and this the fraction would be:

$$\frac{2\pi m_D}{2eB}$$

Does this look more right?

TFM

 

[TFM]

Does the above look correct now?

TFM

 

[michalll]

Yes.

 

[TFM]

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

 

[michalll]

Use equation of equivalence of potential and kinetic energy (potential energy turns to kinetic)

 

[TFM]

Kinetic Energy:

$$K.E. = \frac{1}{2}mv^2$$

Electric Potential:

$$U= q_0V$$

Equate:

$$\frac{1}{2}mv^2 = q_0V$$

$$V = \frac{mv^2}{2q_0}$$

Is this correct?

TFM

 

[TFM]

Putting in my values I get

$$\frac{4 \Pi^2m_D^3}{\frac{4e^2B^2}{2e}}$$

Does this look correct?

TFM

 

[michalll]

Nope, there is no place for pi here v=omega*r.

using omega=eB/m you should get it

 

[TFM]

I looked at the wrong part - i had already found the speed to be:

$$v = \frac{reB}{m_D}$$

so putting in this v, I get:

$$V = \frac{m_D(\frac{r^2e^2B^2}{m_D^2})}{2e}$$

is this better?

TFM

 

[michalll]

This should be correct.

 

[TFM]

Indeed it is correct.

Thanks, michalll

TFM