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

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The discussion revolves around calculating the electric field acting on an electron in the presence of both electric and magnetic fields, using the Lorentz force law. The initial velocity of the electron is given as (12j + 15k) km/s, with a constant acceleration of (2.0 x 10^12 i) m/s^2 and a magnetic field of 400i microTeslas. The user initially calculated the electric field E but neglected the negative charge of the electron, which affects the direction of the force. After correcting for the negative charge, the user arrived at the final expression for the electric field as E = -11.375 i - 6 j + 4.8 k Newtons per coulomb. The discussion highlights the importance of careful consideration of charge signs in electromagnetic calculations.
mfoley14
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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.
 
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mfoley14 said:
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?
 
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?
 
mfoley14 said:
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.
 
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
 
mfoley14 said:
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.
 
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!
 
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