Group of partcles in a magnetic field

[TheWire247]

Homework Statement

A group of particles is traveling in a magnetic field of unknown magnitude and direction. You observe that a proton moving at 1.60 km/s in the + x-direction experiences a force of 2.10×10−16 N in the + y-direction, and an electron moving at 4.30 km/s in the - z-direction experiences a force of 8.50×10−16 N in the +y-direction.

A) What is the magnitude of the magnetic field?

B) What is the direction of the magnetic field? (in the xz-plane) (from the -ve z direction)

C) What is the magnitude of the magnetic force on an electron moving in the - y-direction at 3.40 km/s?

D) What is the direction of this the magnetic force? (in the xz-plane) (from the -ve x direction)

Homework Equations

F=qvxB

The Attempt at a Solution

I tried using the above formula to no success

 

[gabbagabbahey]

Homework Statement

A group of particles is traveling in a magnetic field of unknown magnitude and direction. You observe that a proton moving at 1.60 km/s in the + x-direction experiences a force of 2.10×10−16 N in the + y-direction, and an electron moving at 4.30 km/s in the - z-direction experiences a force of 8.50×10−16 N in the +y-direction.

A) What is the magnitude of the magnetic field?

B) What is the direction of the magnetic field? (in the xz-plane) (from the -ve z direction)

C) What is the magnitude of the magnetic force on an electron moving in the - y-direction at 3.40 km/s?

D) What is the direction of this the magnetic force? (in the xz-plane) (from the -ve x direction)

Homework Equations

F=qvxB

The Attempt at a Solution

I tried using the above formula to no success

You tried; okay, can you show us your attempt?

 

[TheWire247]

F=qvxB

F/q=vxB

F/q=1600i x B

F/q=-1600j + 1600k

(2.1x10^-16/1.6*10^-19)/1600 = B

Then used Pythagoras to calculate the magnitude of B

 

[gabbagabbahey]

F/q=1600i x B

F/q=-1600j + 1600k

Where does this 2nd equation come from? You don't know what $\mathbf{B}$ is; that's what you're supposed to calculate.

What you do know is that $$\mathbf{F}=q_{\text{proton}}(1600\text{km/s})\mathbf{i}\times \mathbf{B}=(-2.1\times 10^{-16}\text{N})\mathbf{j}$$ as well as a similar equation for the force that the electron experiences.

I'd suggest that you let $\mathbf{B}=B_x\mathbf{i}+B_y\mathbf{j}+B_z\mathbf{k}$ and carry out the cross product in both your equations and compare terms to solve for the components of $\mathbf{B}$

 

[asteriatic]

. I also tried using the right hand rule to determine the direction of the magnetic field, but I am unsure of the results.

I would approach this problem by first understanding the basic principles of how particles interact with magnetic fields. The equation F=qvxB is a good starting point, as it relates the force experienced by a charged particle (F) to its charge (q), velocity (v), and the strength and direction of the magnetic field (B).

To find the magnitude of the magnetic field (B), we can rearrange the equation to B=F/(qv). Plugging in the values given in the problem, we get B=(2.10×10−16 N)/(1.602×10−19 C x 1.60 km/s)=8.27×10−5 T. This means that the magnitude of the magnetic field is 8.27x10^-5 Tesla.

To determine the direction of the magnetic field, we can use the right hand rule. If we point our thumb in the direction of the particle's velocity, and our fingers in the direction of the magnetic force, our palm will be facing the direction of the magnetic field. In this case, since the force on the proton is in the +y-direction, and the velocity is in the +x-direction, the magnetic field must be in the +z-direction. For the electron, the force is in the +y-direction and the velocity is in the -z-direction, so the magnetic field must be in the +x-direction.

To find the magnitude of the magnetic force on an electron moving in the -y-direction at 3.40 km/s, we can again use the F=qvxB equation. Plugging in the values, we get F=(8.50×10−16 N)/(1.602×10−19 C x 3.40 km/s)=1.41×10−4 N. This means that the magnetic force on the electron is 1.41x10^-4 Newtons.

Finally, to determine the direction of this magnetic force, we can again use the right hand rule. If we point our thumb in the direction of the particle's velocity and our fingers in the direction of the magnetic field, our palm will face the direction of the magnetic force. Since the velocity is in the -y-direction and the magnetic field is in the +x-direction, the magnetic force must be in the +z-direction.