Group of partcles in a magnetic field

AI Thread Summary
A group of particles is subjected to a magnetic field, with a proton and an electron experiencing forces in the y-direction due to their respective velocities. The magnetic force equation F=qvxB is applied to determine the magnetic field's magnitude and direction. The discussion emphasizes the need to express the magnetic field as a vector and perform cross products to solve for its components. Participants suggest comparing terms from the equations derived for both the proton and electron to find the magnetic field's characteristics. The conversation highlights the complexity of the calculations and the importance of correctly applying the physics principles involved.
TheWire247
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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
 
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TheWire247 said:

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?
 
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
 
TheWire247 said:
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}
 
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