What Forces and Powers Are Involved in Moving a Wire Through a Magnetic Field?

  • Thread starter Thread starter cse63146
  • Start date Start date
  • Tags Tags
    Force Push Wire
Click For Summary

Homework Help Overview

The discussion revolves around a physics problem involving a wire moving through a magnetic field, specifically focusing on the forces and power associated with this motion. The problem includes calculations for pushing force, power input, induced current, and power dissipation in a resistor.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants discuss various equations related to the problem, including those for pushing force and induced current. There are attempts to calculate the induced current and questions about the direction of the induced current using the right-hand rule.

Discussion Status

Some participants have shared their calculations and expressed confusion regarding specific parts of the problem, particularly the direction of the induced current. There is acknowledgment of differing answers for the induced current, with one participant seeking confirmation on their result.

Contextual Notes

Participants are working under the constraints of homework rules, which may limit the sharing of complete solutions. There is also mention of significant figures in calculations, indicating a focus on precision in the answers provided.

cse63146
Messages
435
Reaction score
0

Homework Statement



The 10-cm-wide, zero-resistance slide wire shown in the figure is pushed toward the 2.0\Omegaresistor at a steady speed of 0.50 m/s. The magnetic field strength is 0.50 T.

knight_Figure_33_44.jpg


A) How big is the pushing force?
B) How much power does the pushing force supply to the wire?
C) What is the magnitude of the induced current?
D) How much power is dissipated in the resistor?

Homework Equations



F_{push} = \frac{v L^2 B^2}{R}

P_{input} = F_{push}*v = \frac{v^2 L^2 B^2}{R}

The Attempt at a Solution



A) F_{push} = \frac{v L^2 B^2}{R} = \frac{(0.5) (0.1)^2 (0.5)^2}{2} = 6.25 * 10^{-4} N

B) P_{input} = F_{push}*v = \frac{v^2 L^2 B^2}{R} = \frac{(0.5)^2 (0.1)^2 (0.5)^2}{2} = 3.125 * 10^4 W

C) I = \frac{vLB}{R} = \frac{(0.5)((0.1)(0.5)}{2} = 1.25 * 10^2 A

D) P_{dissipated} = P_{input} = 3.125 * 10^4 W

Did I make any mistakes?
 
Last edited:
Physics news on Phys.org
I am working on exactly same one.

I used I=E/R=(vLB)/R equation to solve C, and had 0.0125A or 1.25*10^-2

but they said it's not quite. :-(

Anyway, there is another question: What is the direction of the induced current?

I know I need to use the right hand rule but can't apply it right.

Is it clockwise or counterclockwise? anyone can explain?
 
The induced current opposes the change in magnetic flux through the loop. As you push the bar to the right you are decreasing the amount of "out of the page" magnetic field so the loop creates a current that would result in MORE magnetic field "out of the page".

Put your hand in the "thumbs up" position and point your thumb in the direction of the magnetic field you want to create...right now, "out of the page". The direction your fingers curl is the direction of the induced current. (ie along your fingers towards your fingernails.
 
man, still confused with "out of the page"

but.. okay. somewhat got it...

Still, nobody for part C ?
 
dmsgo89, I'm pretty sure your answer for C is correct, or is at least the answer I got. Check your significant figures though.
 

Similar threads

  • · Replies 2 ·
Replies
2
Views
959
Replies
4
Views
2K
Replies
3
Views
823
  • · Replies 3 ·
Replies
3
Views
1K
  • · Replies 3 ·
Replies
3
Views
1K
Replies
4
Views
3K
  • · Replies 9 ·
Replies
9
Views
2K
  • · Replies 18 ·
Replies
18
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K