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
SUMMARY

The discussion focuses on the physics of a wire moving through a magnetic field, specifically addressing the calculations of the pushing force, power input, induced current, and power dissipation in a resistor. The pushing force is calculated using the formula Fpush = (v L2 B2) / R, resulting in 6.25 x 10-4 N. The power input is determined to be 3.125 x 104 W, and the induced current is calculated as I = (v L B) / R, yielding 1.25 x 102 A. The discussion also emphasizes the application of the right-hand rule to determine the direction of the induced current.

PREREQUISITES
  • Understanding of electromagnetic induction principles
  • Familiarity with Ohm's Law (I = V/R)
  • Knowledge of the right-hand rule for determining current direction
  • Basic proficiency in using formulas for force and power in physics
NEXT STEPS
  • Study the concept of electromagnetic induction in depth
  • Learn about the applications of the right-hand rule in various scenarios
  • Explore the relationship between magnetic fields and induced currents
  • Investigate the effects of resistance on power dissipation in circuits
USEFUL FOR

Students studying physics, particularly those focusing on electromagnetism, as well as educators and anyone interested in understanding the principles of induced currents and magnetic fields.

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
919
Replies
4
Views
2K
Replies
3
Views
637
  • · 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
1K