Multiloop circuit, magnetic force on a current carrying wire, Ampere's Law

In summary: Similarly the y component of the force involves the z component of the magnetic field.3. In summary, for the first conversation, i1 can be calculated by using the equation i = V/R and solving for I1 using the emf and the resistances R1 and R2. For the second conversation, the force on the wire can be calculated by using the equation for the magnitude of the force on a current-carrying wire and considering the angle between the length vector and the magnetic field components. For the third conversation, the magnitude of the current in the wire can be found by using Ampere's Law and setting the ratio of the magnetic field at point P to the magnetic field at the center of the pipe equal to 3.
  • #1
irnubcake
5
0
1. Multiloop circuit
In Fig. 27-71, R1 = 5.76 , R2 = 18.7 , and the ideal battery has emf = 13.8 V. (a) What is the size of current i1?

http://img253.imageshack.us/img253/9926/fig2771dw3.gif [Broken]

Relevant equations
i = V/R

The attempt at a solution
I redrew the 3 resistors (R2s) as parallel resistors. So i1 would be in the middle. Then all the resistances would have the same voltage 13.8V. So
i1 = V/R2 = 13.8V/18.7 = 1.59 A. It's wrong though.

2. Magnetic Force on a Current-Carrying Wire
A wire 62.9 cm long carries a 0.480 A current in the positive direction of an x-axis through a magnetic field with an x component of zero, a y component of 0.000230 T, and a z component of 0.0120 T. Find the magnitudes of x, y, and z components of the force on the wire.

Relevant equations
Magnitude of the force on a current carrying wire = (current)*(length)*(magnetic field)*(sin (angle between the directions of field and the length))

The attempt at a solution
The length of wire is along the x axis, so the angle between the wire and the x-axis = 0. The angle between the x-axis and y-axis = 90 degrees, between x and z is also 90 degrees:
force magnitude of x component = 0
force magnitude of y component = 0.480A x 0.629m x 0.000230T x sin 90
force magnitude of z component = 0.480A x 0.629m x 0.0120T x sin 90

Only the first one's right.

3. Ampere's Law
In Fig. 29-63, a long circular pipe with outside radius R = 2.65 cm carries a (uniformly distributed) current i = 9.16 mA into the page. A wire runs parallel to the pipe at a distance of 3.00R from center to center. Find the magnitude of the current in the wire in milliamperes such that the ratio of the magnitude of the net magnetic field at point P to the magnitude of the net magnetic field at the center of the pipe is 3.93, but it has the opposite direction.

http://img442.imageshack.us/img442/8181/fig2963ho7.gif [Broken]

Relevant equations
Magnetic field at distance r outside long straight wire with current =
(μ0 x i)/(2 x pi x r)
Magnetic field around Amperian loop = μ0 x current enclosed

The attempt at a solution
(Magnetic Field at pt P) / (Magnetic Field at the center of the pipe) = 3.93

so: (μ0 x i)/(2 x pi x R) = 3.93 x (μ0 x current enclosed)
...i = 3.93 x current enclosed x 2 x pi x R
= 3.93 x 9.16 mA x 2pi x 0.0265m = 0.00599 (mA)(m)
The units should only have mA so the answer is wrong.
 
Last edited by a moderator:
Physics news on Phys.org
  • #2
1. You are corrrect in assuming that the R2s are in parallel and that they will share the same voltage. The emf divides over this parallel combination and R1 though. What you can assume is that the three parallel branches will draw the same current, I1. This means that 3 x I1 will flow through R1. Use the eq.

Emf = V1 + V2

to solve for I1.

2. The components of the magnetic field and the length vector form a triad of vectors where the length vector points along the x-axis.

The direction of the force components are obtained by rotating the length vector towards each of the magnetic field components.

By rotating the length vector towards the y magnetic field component we obtain a force component in the negative z-axis direction. The z component of the force therefore involves the y component of the magnetic field.
 
  • #3


1. In a multiloop circuit, the total current can be found by applying Ohm's law, which states that the current is equal to the voltage divided by the total resistance. In this case, the total resistance can be found by adding the resistances in series or by using the equivalent resistance of parallel resistors. Therefore, the size of current i1 can be calculated as 13.8V/24.46 = 0.564 A.

2. The magnetic force on a current-carrying wire can be calculated using the equation F = I*L*B*sinθ, where I is the current, L is the length of the wire, B is the magnetic field, and θ is the angle between the direction of the wire and the direction of the magnetic field. In this case, the x component of the magnetic field is zero, so the force on the wire in the x direction is also zero. The y component of the force can be calculated as 0.480A * 0.629m * 0.000230T * sin90 = 0.000069 N. The z component of the force can be calculated as 0.480A * 0.629m * 0.0120T * sin90 = 0.0036 N.

3. Ampere's Law states that the magnetic field around a closed loop is equal to the permeability of free space times the current enclosed by the loop. In this problem, the current enclosed by the Amperian loop is equal to the current flowing through the pipe, which is given as 9.16 mA. Therefore, the magnetic field at the center of the pipe can be calculated as μ0 * 9.16 mA = 0.0234 mT. The magnetic field at point P can be calculated using the formula for the magnetic field around a long straight wire with current, which is equal to (μ0 * i)/(2 * pi * r). Setting the ratio of the two fields equal to 3.93 and solving for the current i, we get i = 0.00602 mA. The units in this case should be mA, not (mA)(m).
 

1. What is a multiloop circuit?

A multiloop circuit is a circuit that has multiple loops or paths for the flow of electric current. This means that there are multiple branches or sections in the circuit that allow electricity to flow through them.

2. How is the magnetic force on a current carrying wire determined?

The magnetic force on a current carrying wire can be determined using the right-hand rule, where the thumb points in the direction of the current, the fingers point in the direction of the magnetic field, and the palm indicates the direction of the magnetic force. The magnitude of the force can be calculated using the formula F = ILB, where I is the current, L is the length of the wire, and B is the magnetic field strength.

3. What is Ampere's Law and how is it used in circuits?

Ampere's Law states that the magnetic field around a closed loop is directly proportional to the electric current passing through the loop. This law is used to calculate the magnetic field strength at a specific point in a circuit by integrating the product of the current and the length element along the closed loop.

4. How does changing the number of loops in a circuit affect the magnetic force?

Increasing the number of loops in a circuit will increase the magnetic field strength and therefore increase the magnetic force on a current carrying wire. The magnetic force is directly proportional to the number of loops, so doubling the number of loops will double the magnetic force.

5. Can Ampere's Law be applied to non-circular loops?

Yes, Ampere's Law can be applied to any closed loop, regardless of its shape. As long as the loop is closed and the current passing through it is known, the magnetic field strength at a point can be calculated using Ampere's Law.

Similar threads

  • Introductory Physics Homework Help
Replies
8
Views
353
  • Introductory Physics Homework Help
Replies
7
Views
1K
  • Introductory Physics Homework Help
Replies
7
Views
138
  • Introductory Physics Homework Help
Replies
1
Views
261
  • Introductory Physics Homework Help
Replies
11
Views
2K
  • Introductory Physics Homework Help
Replies
3
Views
948
  • Introductory Physics Homework Help
Replies
1
Views
933
  • Introductory Physics Homework Help
Replies
10
Views
1K
  • Introductory Physics Homework Help
Replies
1
Views
1K
  • Introductory Physics Homework Help
Replies
1
Views
733
Back
Top