PLease answer ALL of the questions correctly in order to be rated 5 stars.I will

Question

PLease answer ALL of the questions correctly in order to be rated 5 stars.I will rated the first person to answer them all crorrectly, not just the first person to post. Show all of your work to better my understanding. Thank you!

2.) Two long, straight wires are separated by 0.12 m. The wires carry currents of 7.0 A in opposite directions, as the drawing indicates.

(a) Find the magnitude of the net magnetic field at the point A.
_____________T

(b) Find the magnitude of the net magnetic field at the point B.
_____________T

3.) The wire in the figure below carries a current of 13 A. Suppose that a second long, straight wire is placed right next to this wire. The current in the second wire is 25 A. Use Amp?re's law to find the magnitude of the magnetic field at a distance of r = 0.72 m from the wires when the currents are in the same direction and in opposite directions.

4.) A square coil and a rectangular coil are each made from the same length of wire. Each contains a single turn. The long sides of the rectangle are twice as long as the short sides. Find the ratio ?square /?rectangle of the maximum torques that these coils experience in the same magnetic field when they contain the same current.

5.) Two pieces of the same wire have the same length. From one piece, a square coil containing a single loop is made. From the other, a circular coil containing a single loop is made. The coils carry different currents. When placed in the same magnetic field with the same orientation, they experience the same torque. What is the ratio Isquare / Icircle of the current in the square coil to the current in the circular coil?

6.) Two circular coils of current-carrying wire have the same magnetic moment. The first coil has a radius of 0.095 m, has 155 turns, and carries a current of 3.8 A. The second coil has 160 turns and carries a current of 8.8 A. What is the radius of the second coil?
______________m

Explanation / Answer

1) B = permeability of the medium * N * I / Ls

where B is the flux density, N is the number of turns, I is the current intensity & Ls is the lenght of the solenoid.
N/Ls = B / I * permeability of the medium = 6 / 2*10^-7 * 1.5*10^2 = 200,000 turns per meter

Note: the permeability was used 2*10^-7 which is the permeability of air.

2)

You have the formulas correct, watch your signs and BRACKETS.
B = ?0/(2?) (Current) / (Perpendicular distance)
Since ?0=4? E -7 Tm/A, we have:
B1 = (4?E-7 Tm/A)(7.5 A)/[2? (0.030 m)] = 5E-5 T
B2 = (4?E-7 Tm/A)(-7.5 A)/[2? (0.150 m)] = -1E-1 T
So BA = B1 + B2 = ?
(It looks like you just left out the square brackets, hence multiplying Pi and 0.03 and 0.15 instead of dividing them.)

For the point B, the two distances are -0.060 m and +0.060 m. Be careful with the signs. Unlike point A, the two components will have the same sign.

3) The magnetic field looping around an infinite wire is:

B = mu0 I / (2 pi r)

If they are going the same direction, add the currents to get I. If they are going the opposite direction, take the difference. They give you r. Look up the permeability of free space mu0. Plugnchug.

4) T = NBIAsin(90) = = NBIA

P = 4x = (L + L/2)2 = 3L = constant
L = 4/3 x
w = 1/2 L = 2/3 x
As =x^2
Ar = Lw = 8/9 x^2

Ts / Tr = As/Ar = 9/8

5) Area of the circle = (4D^2)/ pi

where D is the length of one side of the square. Remember the circumference of the circle is equal to the length of the square's perimeter.


Remember
C = 2pi R
So
R = C/2pi = 4D/2pi = 2D/pi
Therefore
Aci = pi(R^2) =pi(2D/pi)^2 = (4D^2)/ pi

Isq/Ici = 4/pi

6) Magnetic moment NI A.

N - number of turns
I - current
A- area of the circle.

Given N1I1A1 = N2 I2 A2

A2 /A1 = [N1/N2]* [I1/I2] = [152/169]*[3.7/10.1] = 0.33.

(r2/r1)^2 = 0.33

. r2/r1 = 0.57

r2 = 0.57r1 = 0.57*0.086 = 0.049 m

Related Questions

Navigate

• Browse (All)
• Browse P
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.