# Lorentz Force Law Problem

1. Apr 6, 2008

### Duodora

[SOLVED] Lorentz Force Law Problem

Problem and known data:

34.6 An electron travels with velocity 4.9 X 10^6 m/s in the i direction, through a point in space where the magnetic field is .111 T in the j direction. The force on the electron at this point is F=(9.5 x 10^-14 i - 9.5 x 10^-14 k) What is the electric field in each direction?

2. Relevant equations

F = q(E + v x B)

3. Attempt at a solution:

I've been trying to just plug in the component vectors--ie for the "i" direction:

9.5 x 10^-14 = -q (E) + -q(4.9 x 10^6)(0)

which gives: -593750 N/C in the i direction

And I've tried various other combinations (ie using the magnetic field in every direction-- for example in the i direction-- 9.5 x 10^-14 = -q (E) + -q(4.9 x 10^6)(.111) but I'm still not getting the right answer)

Any help would be much appreciated! Thank you so much!

2. Apr 6, 2008

### rohanprabhu

Here, you treat the Electric field and even the forces as scalars. However, electric field is a vector. Start by taking:

$$E = x \hat{i} + y \hat{j} + z \hat{k}$$

Once you've done that, consider all forces as vectors and go about as:

$$9.5 \times 10^{-14} (\hat{i} - \hat{k}) = -1.602 \times 10^{-19} [(x \hat{i} + y \hat{j} + z \hat{k}) + (4.9 \times 10^{6} \times 0.111)(\hat{i} \times \hat{j})]$$

On calculating it all, compare the components and you'll get your value of 'x', 'y' and 'z', which shall give you your vectorial electric field.

Last edited: Apr 6, 2008
3. Apr 7, 2008

### Duodora

Hello,

Thank you for the reply--unfortunately, I'm still really confused.

I understand that the electric field is a vector--what I was trying to do before was split it up into its components--so, first solve for the electric field in the x direction, then the y direction, then the z. Would it be possible to split it up this way?

I'm an idiot about the forces, ugh. I'm still a bit lost as to how to pull it all together--I'm having kind of a hard time visualizing what's going on.

Could you split it up like this, and try to solve for the $$\hat{i}$$ direction?

(9.5 x 10$$^{-14}$$$$\hat{i}$$)= (-1.602 x 10^$$^{-19}$$)(x$$\hat{i}$$ + 4.9 x 10$$^{6}$$) which would yield: -5493009 N/C

Thanks, and I'm sorry for being dense...

4. Apr 7, 2008

### rohanprabhu

Yes, you can. However it defeats the purpose of vectors.

This is why i recommend using vector notation. In this approach, I don't understand why you've taken an addition term of $4.9 \times 10^{6}$. The force caused by the magnetic field is purely in the $i \times j$ direction i.e. the $k$ direction. So, the force due to the magnetic field has NO component in either the i or the j direction and hence your inclusion of the same is wrong. Plus, you've taken the magnetic field as unity.

Here, is a better way of solving it:

Let the electric field be:

$$E = x \hat{i} + y \hat{j} + z \hat{k}$$

then,

$$9.5 \times 10^{-14} (\hat{i} - \hat{k}) = -1.602 \times 10^{-19} [(x \hat{i} + y \hat{j} + z \hat{k}) + (4.9 \times 10^{6} \times 0.111)(\hat{i} \times \hat{j})]$$

Solving this,

$$-5.93 \times 10^{5} (\hat{i} - \hat{k}) = (x \hat{i} + y \hat{j} + (5.43 \times 10^5 + z) \hat{k})$$

Now, compare the the co-effecients of $\hat{i}$, $\hat{j}$, $\hat{k}$, to get:

$$-5.93 \times 10^{5} \hat{i} = x \hat{i}$$

$$5.93 \times 10^{5} \hat{k} = (5.43 \times 10^5 + z) \hat{k}$$

$$y \hat{j} = 0$$

You can now get the Electric field as a vector. The first comparison that we did, i.e. for the 'i' direction, is what you were trying to do by considering a particular direction only.

5. Apr 7, 2008

### Duodora

Ok, I just spend 10 minutes trying to type up this big post as to why I was still confused. But I think I understand it. My handle on vectors apparently is abysmal.

However, for my answer, I just tried -5.94 $$\times 10^{5}$$ $$\hat{i}$$, 0 $$\hat{j}$$, 50,000 $$\hat{k}$$ and still got it wrong. (I've already used up all my answers, but at this point I'd just like to understand how to do the problem...)

I'm obviously still missing something, but I'm really not sure what...

Thank you so much!

6. Apr 7, 2008

### rohanprabhu

As far as i see it.. you have got the answer right? Maybe i (or you) made a calculation mistake somewhere.. even i'm not sure how the answer is wrong. Do you know what the actual answers are?

7. Apr 15, 2008

### Duodora

Sorry it took me so long to reply--unfortunately I don't--our professor has blocked the answers, so there is no way to see what the correct ones are supposed to be :-\

Anyway thank you for all your help--and I'm going to say it was just a quirk with our homework system.

Thanks!