# Some electromagnetism questions

1. Feb 7, 2010

### nunos

1. A battery is connected between A and B of a bar made of semi-conductor of type P. The positive electrode was connected in A and the negative in B. What's the current direction?

A. from A to B
B. in the semiconductor it cannot pass any current
C. it depends on the type of the battery
D. it depends on the emf of the battery
E. from B to A

Last edited: Feb 7, 2010
2. Feb 7, 2010

### espen180

I'll give you a hint: The electric field goes from positive to negative potential. Now, what is the definition of current? Also, how does current flow in a semiconductor?

3. Feb 7, 2010

### nunos

That being said, I guess it's A.

Since it's being applied a corrent from the battery in point A, the positive pole of the semiconductor P, it will increase the potential there forcing some electrons to pass through.

If the semiconductor was of type N, the asnwer would be B, right?

I still have some confusion about this. The sole existance of a semiconductor of type P without the N-semiconductor it's kind of weird, since in most examples I have seen (not too many, though), they are almost coupled together...

4. Feb 7, 2010

### espen180

I'm not so sure about your statement about n-type semiconductors. It's been a while since I studied semiconductors, but doping a semiconductor, whether P or N-doping, should, if anything, increase its conductivity.

I believe the answer is A.

5. Feb 7, 2010

### nunos

Thanks espen 180 for your help. I can imagine how boring it might be for someone who is into physics to help a noob like me with such simple questions. I will keep this topic open if I have any more doubt.

6. Feb 7, 2010

### espen180

Not at all. :) It's actually one of my main pasttimes.

7. Feb 7, 2010

### nunos

Another question:

An alpha particle is formed by 2 protons and 2 neutrons. If this one alpha particle moves with velocity 6.15e5 m/s in a direction perpendicular to a magnetic field |B| = 0.27T, what's the value of the magnetic force in the particle?

So I suppose the formula to use is the Lorentz Formula:

$$\stackrel{\rightarrow}{F} =q(\stackrel{\rightarrow}{E} + \stackrel{\rightarrow}{v} \times \stackrel{\rightarrow}{B})$$

Since the electric field is non-existent and q = 4 times the mass of a proton

$$F = 6.69*10^{-27}*6.15*10^{5}\times 0.27 = 1.1*10^{-21}N$$

Which is wrong, because the correct solution is $$5.3*10^{-14}N$$

I suppose the problem is with the cross product... Any help is greatly appreciated. Thanks.

8. Feb 7, 2010

Anyone?

9. Feb 7, 2010

### espen180

The charge of an alpha particle is 2e, not 4e. (Two protons and two neutrons.)

10. Feb 8, 2010

### Redbelly98

Staff Emeritus
This statement makes no sense, since charge and mass are different types of quantities. espen180 is correct.

In the future, please start a new thread to post a new question.

11. Mar 9, 2010

### DuncanM.

I need some help with my physics ?'s?!
Can anyone help me?

12. Mar 9, 2010

### Redbelly98

Staff Emeritus