Electric field and magnetic field acting simultaneously on a charged particle

In summary, the problem states that 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. To solve for E, the charge of an electron must be subtracted from the equation and the negative sign must be applied to the result.
  • #1
mfoley14
4
0
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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  • #2
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?
 
  • #3
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?
 
  • #4
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.
 
  • #5
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
 
  • #6
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.
 
  • #7
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!
 

FAQ: Electric field and magnetic field acting simultaneously on a charged particle

What is an electric field?

An electric field is a region of space in which an electrically charged particle experiences a force. This force can either attract or repel the particle, depending on the sign of the charge and the direction of the field.

What is a magnetic field?

A magnetic field is a region of space in which a magnetized object or a moving electrically charged particle experiences a force. This force is perpendicular to both the direction of the magnetic field and the velocity of the charged particle.

How do electric and magnetic fields interact with each other?

When an electrically charged particle moves through a magnetic field, it experiences a force perpendicular to both the direction of the field and the particle's velocity. This force is known as the Lorentz force and is the basis for many technological applications, such as electric motors and generators.

What is the difference between electric and magnetic fields?

The main difference between electric and magnetic fields is the type of force they exert on charged particles. Electric fields exert forces parallel to the field, while magnetic fields exert forces perpendicular to the field. Additionally, electric fields are produced by stationary charges, while magnetic fields are produced by moving charges or magnets.

How do electric and magnetic fields affect the motion of a charged particle?

When both an electric and magnetic field are present, the charged particle will experience a combined force that is the vector sum of the two individual forces. The resulting motion of the particle will be a curved path, known as a helix. The exact trajectory of the particle will depend on the strength and orientation of the fields and the initial velocity of the particle.

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