Calculating Magnetic Forces on a Current Loop in a Uniform Field

Click For Summary

Homework Help Overview

The discussion revolves around calculating forces on charged particles and current loops in magnetic and electric fields. The original poster presents a scenario involving an electron moving through electric and magnetic fields, seeking to determine the magnitudes of these fields based on provided data. Additionally, there is a separate inquiry regarding the magnetic forces on a triangular current loop in a uniform magnetic field.

Discussion Character

  • Exploratory, Conceptual clarification, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the relationship between electric and magnetic forces, referencing relevant formulas and graphs to analyze the forces acting on the electron. Questions arise about how to extract values for the electric and magnetic fields from the given graph. In the current loop scenario, there is an exploration of the forces on different sides of the loop and the implications of the calculated values.

Discussion Status

Some participants have provided insights into the relationships between the forces and fields, while others are questioning the calculations related to the current loop. There is an ongoing examination of the assumptions made regarding the forces acting on the loop, with some guidance offered on how to approach the calculations.

Contextual Notes

Participants note the absence of a figure that is crucial for fully understanding the electron's scenario. Additionally, there is a mention of the specific geometry of the current loop and the uniform magnetic field affecting it, which may influence the calculations being discussed.

Jon Wilson
Messages
4
Reaction score
0
i have a question that i need some help with:

at time t1, an electron is sent along the positive direction of an x axis, through both an electric field E and a magnetic field B, with E directed parallel to the y axis. Figure 28-34 gives the y component Fnet, y of the net force of the electron due to the two fields, as a function of the electron's speed v at time t1. the x and z components of the net force are zero at t1. Assuming Bx=0, find the magnitude E and B in unit-vector notion.
 
Physics news on Phys.org
Not enough information given for anyone else to do the problem without the figure. What have you done to try to solve this?
 
there is a graph that has v as the x-axis and Fnet as the y-axis. it is a line that goes from x=0, y=-2 to x=50, y=0 to x=75, y=1. i know in ties into the formula v=E/B and possiably F=(qv)B. what i don't know is how to figure out what B is, if i knew how to do that i would be able to figure out the problem.
 
the best i could come up with is E= -1.25x10^-38 N/C and B= (2.5x10^-2 T) ^k
 
Jon Wilson said:
i have a question that i need some help with:

at time t1, an electron is sent along the positive direction of an x axis, through both an electric field E and a magnetic field B, with E directed parallel to the y axis. Figure 28-34 gives the y component Fnet, y of the net force of the electron due to the two fields, as a function of the electron's speed v at time t1. the x and z components of the net force are zero at t1. Assuming Bx=0, find the magnitude E and B in unit-vector notion.

Jon Wilson said:
there is a graph that has v as the x-axis and Fnet as the y-axis. it is a line that goes from x=0, y=-2 to x=50, y=0 to x=75, y=1. i know in ties into the formula v=E/B and possiably F=(qv)B. what i don't know is how to figure out what B is, if i knew how to do that i would be able to figure out the problem.

You are given that the fields are

[tex]\overrightarrow E = E_y \widehat j[/tex]

[tex]\overrightarrow B = B_y \widehat j + B_z \widehat k[/tex]

You know that the electric force is independent of velocity, so when you look at the graph where the velocity is zero, the net force must be from the electric field alone. You are also told that there is no net force in the x and z directions. What you must do is determine from the direction of the force, and the direction of the motion, which component of the magnetic field is producing the magnetic force. From the force versus velocity graph, you can determine the amount of force being contributed by the magnetic field for any velocity. You can pick one velocity (for example, the one that yields zero net force overall) and find the corresponding magnetic force (remember, the electric force is constant) and use that to determine the strength of the magnetic field.
 
thanks for the explanation. I have one more question for you:

A single-turn current loop, carrying a current of 4 A, is in the shape of a right triangle, with sides 50 cm, 120 cm, and 130 cm. The loop is in a uniform magnetic field of magnitude 75 mT whose direction is parallel to the current in the 130 cm side of the loop. What is the magnitude of the magnetic force on (a)the 130 cm side, (b) the 50 cm side, and (c) the 120 cm side? (d) What is the magnitude of the net force on the loop?

I worked it out and the answers I got were: (a) Fb= 0 (b) .0138 N (c) .0138 N (d) 0. Does this look correct to you?
 
Jon Wilson said:
thanks for the explanation. I have one more question for you:

A single-turn current loop, carrying a current of 4 A, is in the shape of a right triangle, with sides 50 cm, 120 cm, and 130 cm. The loop is in a uniform magnetic field of magnitude 75 mT whose direction is parallel to the current in the 130 cm side of the loop. What is the magnitude of the magnetic force on (a)the 130 cm side, (b) the 50 cm side, and (c) the 120 cm side? (d) What is the magnitude of the net force on the loop?

I worked it out and the answers I got were: (a) Fb= 0 (b) .0138 N (c) .0138 N (d) 0. Does this look correct to you?

a) is correct. b) and c) should not be the same, so d) cannot be right. How are you doing b) and c)?
 

Similar threads

Replies
4
Views
2K
Replies
4
Views
2K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 9 ·
Replies
9
Views
2K
Replies
2
Views
2K
Replies
12
Views
2K
  • · Replies 8 ·
Replies
8
Views
2K
Replies
2
Views
1K
Replies
8
Views
2K
Replies
1
Views
2K