MHB Electric field and magnetic field acting simultaneously on a charged particle

[mfoley14]

The problem that I'm working on states: "An electron has initial velocity (12j +15k)km/s and a constant accelaration of (2.0 x 10^12 i) m/s^2 in a region of uniform electric and magnetic field. If B equals 400i microTeslas, find electric field E.

I know how to do most of the components of this problem, for example a particle traveling through a magnetic field or an electric field. However, I'm not sure where exactly to begin when addressing a particle which has both E and B acting on it.

 

[I like Serena]

The problem that I'm working on states: "An electron has initial velocity (12j +15k)km/s and a constant accelaration of (2.0 x 10^12 i) m/s^2 in a region of uniform electric and magnetic field. If B equals 400i microTeslas, find electric field E.

I know how to do most of the components of this problem, for example a particle traveling through a magnetic field or an electric field. However, I'm not sure where exactly to begin when addressing a particle which has both E and B acting on it.

Hi mfoley14! Welcome to MHB! (Smile)

The Lorentz force in the presence of both an electric and a magnetic field is:
$$\mathbf F =q (\mathbf E + \mathbf v \times \mathbf B)$$
How far can you get with that?

 

[mfoley14]

Using the Lorentz Force Law, I was able to sub out components of the formula with known values. I solved for f using the accelaration and the mass of an electron, substituted the charge of an electron for q, and found the cross product of v and B. So, f = q(E + v x B) became (9.1x10^-31)(2x10^12)i = (1.6x10^-19)(E +6j - 4.8k). I then solved for E and found that E = 11.375 i -6 j + 4.8 k Newtons per coulomb. Is this correct?

 

[I like Serena]

Using the Lorentz Force Law, I was able to sub out components of the formula with known values. I solved for f using the accelaration and the mass of an electron, substituted the charge of an electron for q, and found the cross product of v and B. So, f = q(E + v x B) became (9.1x10^-31)(2x10^12)i = (1.6x10^-19)(E +6j - 4.8k). I then solved for E and found that E = 11.375 i -6 j + 4.8 k Newtons per coulomb. Is this correct?

Almost!
The charge of an electrion is $-1.6\cdot10^{-19}\textrm{ C}$ .
Note the minus sign that effectively reverses the Lorentz force.

 

[mfoley14]

I knew that! Adding the negative sign to my calculation only switches the sign of each component, so the correct answer I believe is E = -11.375 i + 6 j -4.8 k

 

[I like Serena]

I knew that!

Good!

Adding the negative sign to my calculation only switches the sign of each component, so the correct answer I believe is E = -11.375 i + 6 j -4.8 k

Erm... it only switches the sign of the x component.

 

[mfoley14]

Wow that was some lazy math on my part. I spotted my mistake. E = -11.375 i - 6 j + 4.8 k.
Thank you for your help!