# Solenoid Investigation

Extracts from this document...

Introduction

Oxford Cambridge and RSA Examinations

Advanced Subsidiary GCE

PHYSICS A (2823/03/PLAN)

Practical Test (Part A - Planning Exercise)

Research/Theory

“The moment of a (turning) force about a given pivot is defined as follows.

Moment of a force = force x perpendicular distance from the pivot to the line of action of the force. The unit of moment is the Newton metre (Nm).”1

Also, “The principle of moments states that for any body in equilibrium, the sum of the clockwise moments about any pivot must equal the sum of the anticlockwise moments about that pivot.”1(Fig. 1)

“Inside a solenoid, the magnetic field is uniform and its strength can be increased by increasing the current.”2

## The principle of moments is used to find attractive force between the magnet and the solenoid.

The unknown force is the one produced with the bar magnet and solenoid. By using the equation of equilibrium previously given, the unknown force can be calculated provided that the system is in equilibrium:

F1 x d1 = F2 x d2

F1 = (F2 x d2)

Middle

#### Preliminary Investigation

The first aspect that was investigated was the direction of the current (Fig.5). It must be noted that “There is an attractive force”.

The magnitude of the current that will be used through the solenoid was then investigated. When the resistance of the variable resistor was at its smallest the current through the solenoid was 3.42A with a 2V supply (a 2V supply was found adequate as higher e.m.f. was found to trip the power supply). However, at this current flow, it was found that the solenoid would heat up very quickly and to a high temperature. By adjusting the variable resistor to a higher resistance it was found that when set so the current on the ammeter was 0.5A, the solenoid did not heat too much or heat too quickly therefore the resistance of the solenoid would not be affected greatly.

The only meter used was an ammeter. It is

Conclusion

#### Prediction

#### Results Table

- F2 can be calculated by m x 9.81 = F2.
- F1 can be calculated by (F2 x d2) / d1 = F1.

Reading 1

Distance, x/ m | Mass, m/ grams | Force, F2/ Newtons | Force, F1/ Newtons |

0.00 | |||

0.02 | |||

0.04 | |||

0.06 | |||

0.08 | |||

0.10 | |||

0.12 |

Reading 2

Distance, x/ m | Mass, m/ grams | Force, F2/ Newtons | Force, F1/ Newtons |

0.00 | |||

0.02 | |||

0.04 | |||

0.06 | |||

0.08 | |||

0.10 | |||

0.12 |

Reading 3

Distance, x/ m | Mass, m/ grams | Force, F2/ Newtons | Force, F1/ Newtons |

0.00 | |||

0.02 | |||

0.04 | |||

0.06 | |||

0.08 | |||

0.10 | |||

0.12 |

Average

Distance, x/ m | Mass, m/ grams | Force, F2/ Newtons | Force, F1/ Newtons |

0.00 | |||

0.02 | |||

0.04 | |||

0.06 | |||

0.08 | |||

0.10 | |||

0.12 |

#### Bibliography

- New Understanding Physics For Advanced Level by Jim Breithaupt (Stanley Thornes).
- Physics 1 by David Sang, Keith Gibbs & Robert Hutchings (Cambridge OCR).

This student written piece of work is one of many that can be found in our AS and A Level Mechanics & Radioactivity section.

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