O n l i n e E xp e r i m e n t - M a g n e t i c F i e l d

PHYSICS DEPARTMENT

PHYS 192 - Physics II Labs

Online Experiment - Magnetic Field in Circular Coil

Line 1: Name of the experiment? (Magnetic Field in Circular Coil)

Line 2: Your full name: (last, first)

Line 3: Lab course? (Phys 192)

Line 4: Section? (two digits code)

Line 5: Purpose of the experiment? (find the purpose after reading the theory and procedures)

Theory

In this lab we will be concerned with measuring how the magnetic field is sourced by currents. The magnetic analogue of Coulomb’s law is named Biot and Savart law. For an infinitesimal source at the origin, the infinitesimal contribution to the field is given by

Biot-Savart: B d→

= μ04π r2

Idl×r→

ˆ magnetic field at position due to current in a segment r→ I ld→

Coulomb: E d→

= 14πε0 r2

dq r̂electric field at from a charge r→ qd

There are a few key differences between the magnetic and electric cases: the permittivity has been replaced by a different constant ε0

, which is called the permeability of the vacuum and appears in the numerator instead of the denominator; the magnetic π 0 T A μ0 = 4 × 1 7− · m/ field is sourced by current (moving charge) instead of static charge; and the cross product is used to combine the vectors and into a single dlI

→ r̂

vector. This last property means that the geometry of magnetic field lines looks very different from electric field lines, and they tend to wrap

Prepared by the lab supervisor, Dr. A�lgan (ea�lgan01@manha�an.edu) Online 192-Magne�c Field in Circular Coil PAGE 1 / 9

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around the source current, rather than travelling inward or outward as electric field lines do. To find the magnetic at a point, we need to take an B→

integral to combine the contributions coming from all current-carrying segments. Since the current needs somewhere to go, there is no such thing as an isolated current segment, just as there is no such thing as an isolated magnetic point charge.

There are a few current distributions for which we can integrate analytically and only along a special direction that the existing symmetry yields some simplifications. In this lab we will analyze a circular coil.

Circular Coil System

Consider a circular coil of radius carrying a current : we would like to find the a I field at a point a distance along the perpendicular axis. In the figure shown, the x current circles counterclockwise.

The field sourced by a segment has a magnitude given by Bd→

l dφd = a

Bd = μ04π R2

Iadφ = μ04π

Iadφa +x2 2

and is directed as shown. Here is the angle around the circle (not shown in the φ figure).

We can integrate over the circumference of the ring to find the total magnetic field. All components will cancel out except for the x-component, which is

. B sin(θ)d x = μ04π

Iadφa +x2 2 = μ0

4πIadφa +x2 2

a(a +x )2 2 1 2/


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. Bx = μ04π ∫

2π

0

I a dφ2

(a +x )2 2 3 2/ = μ a I02

2(a +x )2 2 3 2/

The final expression should be multiplied by the number of times the wire is wound around the coil. N

Bx = μ a NI02

2(a +x )2 2 3 2/

Note that , and along the x axis. By = 0 Bz = 0 a )Bx ~ ( 2 + x2 3 2− /


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Procedure

In this online lab, you will justify that is proportional . Bx a )( 2 + x2 3 2− /

In the theory section it has been shown that Bx = μ a NI02

2(a +x )2 2 3 2/ where is the radius of the coil and is the distance to the center of the a x coil.

We will measure the magnetic field along the symmetry axis of a coil (the x axis) at points that are at a distance to the center of the coil as the multiples of the radius of the coil, meaning where n x = a , , , .. n = 1 2 3 .Note that is different than which is the number of loops rounding n N the coil (default value of N is 4 in the simulation).

If we incorporate this information (i.e. ) into our formula: n x = a (1 )Bx = μ a NI02

2(a +a n )2 2 2 3 2/ = μ NI0

2a (1+n ) 2 3 2/ = 2a μ NI0 + n2 3 2− /

Let us take the logarithm of both sides: . (B ) ( ) (1 ) ln x = ln 2a μ NI0 − 2

3 ln + n2

In this form, is linearly related to . (B ) ln x (1 ) ln + n2

Line 6: The slope of this linear relationship is _________ . Hint: Consider that if and then . (1 ) x = ln + n2 (B ) y = ln x xy = m + b


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Step 1: Position the coil on the grid. Go to https://phet.colorado.edu/en/simulation/legacy/magnets-and-electromagnets . Download and run the java file you downloaded. You may have to install java if you do not have it on your computer. Google how to run java files.

You will see this initial window show up when you run the program >>>

Feel free to play with the bar magnet but our experimental setup is on the second tab: “Electromagnet”.

<<< Click the tab, you will see this.


https://phet.colorado.edu/en/simulation/legacy/magnets-and-electromagnets

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Get rid of the big compass (unclick “Show Compass”) and show the magnetic field sensor. Click “Show Field Meter”: You will see this >>>

<<< Position the coil and the sensor as shown below: put them near each other, slightly on the left side from the central region.

Note that the little compasses on the background are oriented in the direction of magnetic field lines and they are positioned regularly on a grid. The distance between the grid points (i.e. the little magnets) is the same with the radius of the coil as you can check this visually (this statement holds in approximate terms).


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Now expand the window to the full screen in order to take advantage of the best resolution, and position the sensor at the center of the little magnet. >>>

<<< Next, move the coil near the sensor in both x and y directions until you satisfy both following conditions below:

99 280 < Bx < 2 and

. 0 B 0 − 2 < y < 2

Please note that should NOT be 300 G and the magnetic Bx field sensor measures in units of Gauss (i.e. G). In SI units the unit of the magnetic field is Tesla T and . T 0 G1 = 1 4

If you achieved this step then you have just passed the most difficult part of the experiment! NEVER MOVE THE COIL AGAIN.


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Step 2: Measure along the x axis. Now you are Bx ready to measure the x component of the magnetic field along the x axis at points where Bx a x = n n=1, 2, …, 8. >>>

Please do so and record the values of the magnetic field. The cross of the sensor should be positioned at the center of the little magnets.

Step 3: Process the data: Open a google spreadsheets program. On the first column enter n=1,2,...,8 and title the column as “n”. Enter in units Bx of Gauss into the second column and title is as “B [G]”. On the third column convert the magnetic field into tesla and title the column as “B [T]”. On the fourth column compute and title it as “ln(1+n^2)”. On the fifth column compute by using the third column values not the (1 ) ln + n2 (B) ln second column and title it as “ln(B)”. Logarithms do not have physical units! Plot ln(B) versus ln(1+n^2), do the line fit and find the slope and the intercept:

Line 7: The numerical value of the slope is ____________ .

Line 8: The unit of the slope is ___________ .

Line 9: How does the slope compare to the theoretical expectation (Line 6)? Give your answer as the percent discrepancy between the two.

Line 10: What is the value of the intercept?

Line 11: What is the unit of the intercept?

Line 12: The theoretical expectation of the intercept is , see page 4, where N=4 ( the default parameter in the simulation). (μ NI 2a) ln 0 / Find the current if the radius is given as . Give your answer in units of kilo amps. Take in SI units. I 5 cm a = 2 4π×10μ0 = 7−

Line 13: State your conclusion.


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Your spreadsheet work should look like this:


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