What is the direction of the magnetic force on a charged particle?

[4Phreal]

1. Here is the prompt:

http://imgur.com/mfbPidG

2. F = qv x B
3. At first this seemed like a simple cross product problem, and it probably still is, but I'm really confused as to what "3.70E6 m/s/ in the (i+j+k)/sqrt(3) direction" means, so I don't know how to set up my problem anymore. Could someone instruct me?

 

[berkeman]

1. Here is the prompt:

http://imgur.com/mfbPidG




2. F = qv x B



3. At first this seemed like a simple cross product problem, and it probably still is, but I'm really confused as to what "3.70E6 m/s/ in the (i+j+k)/sqrt(3) direction" means, so I don't know how to set up my problem anymore. Could someone instruct me?

You are given v as a vector and B as a vector, both in rectangular coordinates. Just write the equation for the cross product, and solve for the components of F...

 

[4Phreal]

Right, um, but I don't know what the phrase in the problem that states the velocity vectors actually means (because I'm dumb), so I don't know what those are. Could you please explain what that phrase means?


Here is my attempt: http://imgur.com/5C7Z648

 

[berkeman]

Right, um, but I don't know what the phrase giving me the vectors for v means (because I'm dumb), so I don't know what those are. Could you please explain what that phrase means?

v is the velocity magnitude multiplied by that unit vector u that you are given.

Are you familiar with how to use a Determinant to do the vector cross product in rectangular coordinates?

http://en.wikipedia.org/wiki/Cross_product

.

 

[4Phreal]

I'm familiar, I just don't know how to turn (i+j+k)/sqrt(3) into i, j, and k individually to perform the cross product. Is my attempt in the previous post how I would do it?

 

[berkeman]

Right, um, but I don't know what the phrase in the problem that states the velocity vectors actually means (because I'm dumb), so I don't know what those are. Could you please explain what that phrase means?Here is my attempt: http://imgur.com/5C7Z648

I'm familiar, I just don't know how to turn (i+j+k)/sqrt(3) into i, j, and k individually to perform the cross product. Is my attempt in the previous post how I would do it?

That's close, but has a couple errors in it. First, the velocities are 3.7E6/√3, not 3.7E6/3√3. Where did that extra 3 in the denominator come from? You just distribute the 3.7E6 across each of the unit vectors to get the individual components. Does that make sense?

And in your Determinant calculation, you correctly show the two terms subtracting first, and then in the next line you show them adding...

 

[4Phreal]

I got the second 3 because I thought that (^i + ^j + ^k) divided by sqrt(3) equalled the velocity, so to get each individual part I divided by 3. But what you're saying is that they each individually equal 3.70E6 divided by sqrt(3), so there's no need to divide by that 3?

In my determinant calculation, the Bz is equal to -8.29, changing the subtraction sign into an addition one.

 

[berkeman]

I got the second 3 because I thought that (^i + ^j + ^k) divided by sqrt(3) equalled the velocity, so to get each individual part I divided by 3. But what you're saying is that they each individually equal 3.70E6 divided by sqrt(3), so there's no need to divide by that 3?

In my determinant calculation, the Bz is equal to -8.29, changing the subtraction sign into an addition one.

Ah, I see now what you did with the sign.

Yes, to get the velocity, you just distribute the amplitude through with multiplication. No other operation is needed. Just like 5(i + j + k) = 5i + 5j + 5k.

 

[4Phreal]

Thanks! For some reason, the answer was positive instead of negative, but maybe that was just in the way the question was phrased.