EM11 MAGNETIC FIELD IN A SOLENOID HALL EFFECT PROBE

OBJECTIVES

To become familiar with the use of the Hall-Effect Device for measuring magnetic fields; to verify the relationship between the current in the windings of a solenoid and the magnetic field it produces at the center of the solenoid; to explore the variation of the magnetic field within the solenoid core as a function of distance inside the core for a constant solenoid windings current.

To observe the effect that different materials in the solenoid core have on the magnetic field values.

PROBLEM ASSIGNMENTS

1. An air-core solenoid 25 cm long has 1500 turns. The windings of the solenoid carry a current of 1.15 Amp. Compute the magnetic field at the center of the solenoid core. Assume that the formula for an infinitely long solenoid applies. Answer: B = 8.7 x 10-3 Tesla

2. The transverse voltage (VH) developed across a Hall-Effect Device is proportional to the external B field to which it is exposed. A calibration chart states that VH = KB. The value of K is given as: K = 9.7 Compute the following:

3. Review the Hall-Effect phenomenon and identify conditions under which the Hall-Voltage is not likely to be linearly related to the magnetic field B.

Answer: Changing of drift velocity and/or the number of charge carriers/m3.

4. How many turns are required for an air-core solenoid approximately 50 cm long so as to produce a magnetic field of 55 gauss at the center when a current of 650 mA is flowing in the windings?

Answer: 3367 turns

APPARATUS

PROCEDURE

Part A: Magnetic field produced at the center of the solenoid by applied current to the solenoid.

Figure 1

Hall Effect Device Probe (HED-P)

Figure 2

Part B: Variation of B field within the solenoid for a fixed current in its windings.

Part C: The effect of different materials in the solenoid core on the B field produced by a fixed current in the windings.

Part A: Current in the windings of the solenoid and the resulting B field at its center. HED-Probe # Probe Sensitivity, K = Volts/Tesla , # of turns in coil, N = Length of Coil, L = ___________

Table #1: B field in the center of the solenoid as a function of varying currents in the windings.

Obs. Current (I) (mA) Sensor Voltage (VH) ( Volts ) B ( Tesla ) 1 0.0 2 200 3 400 4 600 5 800 6 1000 7 1200 8 1400 9 1600 10 1800

Sample calculation of B field for Table #1 and Table #2:

Given VH = KB ; B = = (Record result below)

For current = , K = , B =

Part B: Variation of B field within the core of the solenoid for a fixed current in the windings.

Table #2: Variation of B field with an air core solenoid as a function of location

Steady Current =

Location X ( cm ) VH ( Volts ) B ( Gauss ) 0 1.5 3.0 4.5 6.0 7.5 9.0 10.5 12.0 13.5 15.0

Part C: Effect of different materials in the solenoid core on the B field produced by a steady winding current.

Table #3: Bmax in the core in the presence of different materials.

Current in the solenoid = .

Material VH (max) B (max) Air Aluminum Copper Steel

Observations arising from steps r and s in Part C.

Report : For this lab you are required to write a conclusion/discussion section for your results in the usual format with the following points in mind. (Do not forget to show any relevant calculations or data analysis in its own section)

Part A

Part B

Part C

Appendix 1

B FIELD IN A SOLENOID

THEORY

Refer to your textbook for the following topics: B field at the center of a circular current carrying conductor and/or the B field of a solenoid. You will find that the B field at the center of an infinitely long solenoid (which essentially means high ratio of length of solenoid in relation to its diameter) is given by:

B = µo(n)I = µo(N/ )I (1)

where µo = mag. permeability of free space = 4 x 10-7

n = turns per meter (m-1);

=length of coil (m)

N = total turns in length of coil = (n) x ( )

I = Current (Amp)

B = magnetic field in Tesla

(Recall that 1 Tesla = 10,000 Gauss.)

If the solenoid is not infinitely long, then the B field at the center is smaller than that given in Equation 1. For the "finite" solenoid you are using in this lab, the correction factor is calculated to be 0.98. This means that the field at the center of your lab solenoid is 98% of the value it will have if it was an infinitely long solenoid. The Equation 1 for the solenoid you are using becomes:

= (0.98) µo(n) Where = slope of Graph #1(2)

NOTE: Do not confuse the slope constant "B" with the magnetic field symbol B

The output voltage (VH = KB) (3)

Where K = conversion factor, the value of K will be supplied to you in the lab. This value of K may vary with the HED-P used

B =

Where B – magnetic field (T) Vout – output voltage of Hall effect sensor (V) VoQ – output voltage of Hall effect sensor when B = 0 (V) Sens – sensitivity of sensor is different for each device and will be supplied by your instructor Appendix 1