Magnetic field vector using F = qV * B

[happyparticle]

Homework Statement

Magnetic field

Relevant Equations

F = qV * B.

Hi,
I'm trying to find the magnetic field B using F = qV * B.
I have F = (3i + j + 2k) N
V = (-i +3j) * 10^6 m/s
q = -2 *10^6 C
Bx = 0

I don't know how to resolve a 3 dimensional vector equation.
B = F/qV makes not sense for me.

 

[TSny]

Work with the components of F. That is, write out individual equations for F_x, F_y, and F_z.

Is the factor of 10⁶ correct for q? That's a LOT of charge

 

[happyparticle]

Work with the components of F. That is, write out individual equations for F_x, F_y, and F_z.

Is the factor of 10⁶ correct for q? That's a LOT of charge

10^-6, my bad.

 

[happyparticle]

qV = (2i -6j)

(2i -6j) * B = (3i + j + 2k)I tried something, but I don't if it is the right way.

2 0 = 3
-6 j = 1
0 k = 2
then,
-6k = 3
-2k = 1
2j = 2

k =-0.5 and j = 1
B = `(j - 0.5k) T

 

[TSny]

I agree with your final result, but I can't follow your work.

For example, you wrote:

2 0 = 3
-6 j = 1
0 k = 2

On the left of the first equation, you wrote 2 0. What does this mean?

On the left of the second equation, you wrote - 6 j. Is this -6 multiplied by the unit vector j? How can that equal the right hand side, which is equal to 1?

 

[happyparticle]

It's quite impossible to type it. Basically, it's like a matrix 2i 0i | 3
-6i j | 1
0i k| 2

 

[TSny]

It's quite impossible to type it. Basically, it's like a matrix 2i 0i | 3
-6i j | 1
0i k| 2

I still can't follow this. You got the correct answer, so I think you are probably thinking about it correctly. I just can't follow the way you are writing it.

I was suggesting that you write an equation for just the x-component of the vector equation $\vec F = q \vec v \times \vec B$.

Thus,

F_x = ...

where the right-hand side would be expressed in terms of q and certain components of v and B.

 

[happyparticle]

I'm not sure to understand. Fx = 3i
3i = 2i ?

 

[TSny]

I'm not sure to understand. Fx = 3i
3i = 2i ?

When you take the cross product of two vectors, $\vec b \times \vec c$, the x-component of the cross product is

$(\vec b \times \vec c )_x = b_yc_z -b_zc_y$.

Similarly for the y and z components. See here. This pattern is worth memorizing!

Use this to write out the x-component of the vector equation $\vec F = q \vec V \times \vec B$. That is, write out the right-hand side of

$F_x = q (\vec V \times \vec B)_x$

 

[Orodruin]

You cannot solve for $\vec B$ from the magnetic force equation unless you know the force for more than one velocity. The cross product contains information only about the components of $\vec B$ that are perpendicular to $\vec v$. However, you have additional information as you know that $B_x = 0$.

a) What does this tell you about $\vec B$?
b) what do you then get if you take the inner product of $\vec F$ and $\vec B$?