Some physics work, show me what i am doing wrong.

  • Thread starter Thread starter Pepsi24chevy
  • Start date Start date
  • Tags Tags
    Physics Work
Click For Summary
SUMMARY

The discussion focuses on solving problems related to magnetic fields and forces in physics, specifically involving current-carrying wires and loops. The first problem involves calculating the net force on a rectangular loop due to a straight wire using the formula \( \frac{\mu_0 I_1 I_2}{2\pi} \ln(1 + \frac{a}{c}) \), where the user incorrectly computed the magnitude. The second problem pertains to determining the magnetic field in a coaxial cable using \( \frac{\mu_0 I}{2\pi r} \), which also yielded incorrect results. The conversation highlights the necessity of correctly applying Ampere's circuital law and understanding the direction of currents in multiple conductors.

PREREQUISITES
  • Understanding of Ampere's circuital law
  • Familiarity with magnetic field equations, specifically \( B = \frac{\mu_0 I}{2\pi r} \)
  • Knowledge of electric flux and its relation to changing electric fields
  • Basic principles of electromagnetism, including forces on current-carrying conductors
NEXT STEPS
  • Review the derivation and applications of Ampere's circuital law
  • Study the concept of electric flux and its calculation between capacitor plates
  • Explore the effects of multiple currents on magnetic fields in coaxial cables
  • Practice solving problems involving magnetic forces on loops and wires
USEFUL FOR

Students and educators in physics, particularly those focusing on electromagnetism, as well as anyone preparing for exams involving magnetic fields and forces in current-carrying conductors.

Pepsi24chevy
Messages
65
Reaction score
0
Hey guys, here is the pic for one problem. http://www.webassign.net/pse/p30-17.gif
the question reads:In Figure P30.17, the current in the long, straight wire is I1 = 9.00 A and the wire lies in the plane of the rectangular loop, which carries the current I2 = 10.0 A. The dimensions are c = 0.100 m, a = 0.150 m, and l = 0.550 m. Find the magnitude and direction of the net force exerted on the loop by the magnetic field created by the wire.
I know that the direction is to the left. As for the magnitude i tried to use (munot(I1)(I2)/(2pi)) * ln(1+a/c). when i plug in my numbers i get 1.0363e-4 which is not correct. What am i doing wrong.


next, i have this problem:
http://www.webassign.net/pse/p30-21alt.gif
Figure P30.23 is a cross-sectional view of a coaxial cable. The center conductor is surrounded by a rubber layer, which is surrounded by an outer conductor, which is surrounded by another rubber layer. In a particular application, the current in the inner conductor is I1 = 1.06 A out of the monitor, and the current in the outer conductor is I2 = 3.14 A into the monitor. Determine the magnitude and direction of the magnetic field at point a.
Determine the magnitude and direction of the magnetic field at point b.

I tried to use the equation (munot*I)/(2pi*r) but i get 1.06 which is not correct.
For magnetic field at b, i tried to use((munot*I)/(2piR^2))*r which gets me to 6.98 but is not correct.



Then for this 3rd problem it reads:
A 1.00 A current is charging a capacitor that has square plates, 2.00 cm on a side.
(a) If the plate separation is 6.00 mm, find the time rate of change of electric flux between the plates.

I don't know what to do for this one, so any equation or hint someone can give me would be great, thanks.

For this 4th problem:The magnetic field 41.0 cm away from a long, straight wire carrying current 9.00 A is 4.39 µT.
(a) At what distance is it 0.800 µT?
cm
(b) At one instant, the two conductors in a long household extension cord carry equal 9.00 A currents in opposite directions. The two wires are 3.00 mm apart. Find the magnetic field 41.0 cm away from the middle of the straight cord, in the plane of the two wires.

Your answer differs from the correct answer by orders of magnitude. nT
(c) At what distance is it one tenth as large?
cm
(d) The center wire in a coaxial cable carries current 9.00 A in one direction and the sheath around it carries current 9.00 A in the opposite direction. What magnetic field does the cable create at points outside?
nT


I know A is 225 cm by usingb(2pi*r)= munot Iinc, in which i found Iinc then used that and the new value of b into the same equation in which i got 225 which according to webassigned is correct.
I don't know what to do for b,c,and d. I was told to use b= (munot8I/2pi)(1/r1-1/r2) but i can not get the correct answer.
 
Physics news on Phys.org
1
I think you have to use formulae for magnetic field

B = \frac{\mu _0 I_1}{2 \pi d}

for the magnetic field and then the magnitude of force d = c and c + a

F = BI_2 l

for the two wires separately

the current in both wires is in opposite direction and therefore the resultant force is the difference of the two.
 
2
For part (a) it appears that you have used correct formula. r = 1mm and I = I1.

For part (b) again the same only I = I1 -I2 and r = 3 mm

Go through Ampere's circuital law again.
 

Similar threads

  • · Replies 23 ·
Replies
23
Views
5K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 9 ·
Replies
9
Views
3K
  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 1 ·
Replies
1
Views
4K
  • · Replies 5 ·
Replies
5
Views
1K
Replies
1
Views
1K
  • · Replies 15 ·
Replies
15
Views
6K
  • · Replies 9 ·
Replies
9
Views
13K