COOPER CITY HIGH SCHOOL

AP PHYSICS

DR.RIGOL

 

DETERMINATION OF THE MAGNETIC FIELD INSIDE A HELMHOLTZ COIL

 

INTRODUCTION :

 

The Helmholtz coils is a combination of two circular coils of equal radius that are mounted in parallel along a common axis. The condition for these coils is that the radius of the coils is equal to the distance between the coils.

http://t0.gstatic.com/images?q=tbn:ANd9GcSvXF8x60fzLibsnQBV477omgZWX3BFk2El8VYYHN_U-RLt4G3adw

 

We made the Helmholtz coils using enameled wire with diameter 0.9 mm. The 60 loops in each coil were rolled with a radius 0.059 m.

 

HelmholtzCoilsA.jpg

 

To determine the magnetic field inside the Helmholtz coil, we determine the force acting on a wire conducting a current located inside the magnetic field created by the Helmholtz coils. The conducting current wire is hanging from one arm of a balance. The balance is calibrated to null when no current is going through the wire. When a current is present, the Lorentz force acting on the wire will alter the state of equilibrium in the balance. To restore the point of equilibrium some additional weight is added to the right arm of the balance. The added amount of weight will be equal to the magnitude of the magnetic force acting on the wire. Knowing the magnetic force and the current going through the wire, the Lorentz force formula can be used to calculate the magnitude of the magnetic field. This magnitude of the magnetic field determined in the experiment, is then compared to the calculated value using the Biot-Savart Law.

 

MAGNETIC FORCE ACTING ON A WIRE WITH ELECTRIC CURRENT:

When a wire with length L inside a magnetic field with magnetic induction B is conducting an electric current I, there will be a force F acting on the wire. This force, known as Lorentz’s force, has the following expression:

In this expression L is a vector that has the direction of the current and the magnitude of the wire length. As a consequence of the vector product, the direction of the force will be perpendicular to the direction of the current, and to the direction of the magnetic field.

HELMHOLTZ COILS:

Now, I will derive the expression of the magnetic field created on the axes of symmetry at the middle point between the two coils. I will start with the Biot-Savart law that tells us that the magnetic field created by an element of current I at the distance r is given by the expression:

d

 

In this expression, 0 is the permeability of the vacuum [0 =  (T.m/A)], and   is the element of current creating the magnetic field at the distance.

The element of magnetic field created by Idl is shown in the fig. above. This vector can be represented by one x-component that will be looking to the left, and a y-component that will be looking down. When we integrate around the coil, the final sum of all y-components will be zero and the total magnetic field will be the x-component that will be looking to the left. That is why a factor cos(j) will appear in the above formula.

dB

the relationship between r and R can be found from the graph:

r =                          

The integral of dl will be 2πR, so the value for the magnitude magnetic field B will be:

B = [4/(125)1/2] µ0 I/R 

If we consider that the coil has N loops, and that we have two coils, then the final result will be:

B = (4/5)3/2  µ0 NI/R 

EXPERIMENTAL VALUES FOR B:

Our Helmholtz coils have 60 loops and the radius is R=0.059 m, so the expected value for B will be given by the expression:

Be= 9.1×10-4 I (T)

To find the observed value of B (Bo), we use the expression of the Lorentz force:

Bo =

To increase the effect of the magnetic force, instead of using a simple wire we use a coil with 40 loops. This coil is shaped as a square with mean side size 0.070±0.005 m.

This coil is hanging from the balance and inside the Helmholtz coil in such a way that the top is outside and the bottom is in the center of the Helmholtz coils. Observe that the current going through this coil (Ic)is, in general, different from the current going through the Helmholtz coils (IHC).

#

IHC(A)

IC(A)

F(N)

Bo(mT)=F/(Nc×IC*L)

Be(mT)=9.1×10-4×IHC

[Bo- Be](mT)

1

2

3

0.014

1.7

1.8

-0.1

2

2

4

0.021

1.9

1.8

0.1

3

2

5

0.026

1.9

1.8

0.1

4

3

3

0.024

2.9

2.7

0.2

5

3

4

0.031

2.8

2.7

0.1

6

3

5

0.036

2.6

2.7

-0.1

7

4

3

0.028

3.3

3.6

-0.3

8

4

4

0.038

3.4

3.6

-0.2

9

4

5

0.048

3.4

3.6

-0.2

10

5

5

0.062

4.4

4.6

-0.2