1.An electron gun fires electrons into a magnetic field directed straight downward. Find the direction of the force exerted by the field on an electron for each of the following directions of the electron’s velocity: (a) horizontal and due north; (b) horizontal and 30° west of north; (c) due north, but at 30° below the horizontal; (d) straight upward.

2.(a) Find the direction of the force on a proton moving through the magnetic fields in Figure 1, as shown. (b) Repeat part (a), assuming the moving particle is an electron. [pic]
Figure 1

3.Find the direction of the magnetic field acting on the positively charged particle moving in the various situations shown in Figure 2 if the direction of the magnetic force acting on it is as indicated.

[pic] Figure 2

4.The magnetic field of the Earth at a certain location is directed vertically downward and has a magnitude of 50.0 μT. A proton is moving horizontally toward the west in this field with a speed of 6.20 × 106 m/s. What are the direction and magnitude of the magnetic force the field exerts on the proton?

Magnetic Force on a Current-Carrying Conductor

5.A current I = 15 A is directed along the positive x-axis and perpendicular to a magnetic field. A magnetic force per unit length of 0.12 N/m acts on the conductor in the negative y-direction. Calculate the magnitude and direction of the magnetic field in the region through which the current passes.

6.In Figure 1, assume that in each case the velocity vector shown is replaced with a wire carrying a current in the direction of the velocity vector. For each case, find the direction of the magnetic force acting on the wire.

7.In Figure 2, assume that in each case the velocity vector shown is replaced with a wire carrying a current in the direction of the velocity vector. For each case, find the direction of the magnetic field that will produce the...

...Lab V, Problem 8: MagneticForce on a Moving Charge
John Greavu
Partners: Hannah Eshenaur and David Sturg
August 15, 2013
Physics 1302W, Professor: John Capriotti, TA: Barun Dhar
Objective
In order to improve the design of an electron microscope, we attempted to use a magnetic field to control the electron beam. Utilizing a cathode cay tube (CRT), we oriented a magnetic field perpendicular to the axis of the CRT’s electron beam. From previous experience, we decided on using Helmholtz coils to produce a reasonably uniform magnetic field. We measured the deflection of an electron, (our dependent, “response”, variable), as a function of the magnitude of the magnetic field, , and the electron’s velocity, (our independent, “control”, variables).
Prediction and Procedure
For this experiment, we used the CRT, a digital multimeter (DMM), compass, meter stick, 200-turn (each) Helmholtz coils, a Hall probe to measure the magnetic field (and accompanying computer data acquisition system), and banana cables to make the appropriate connections.
Figure 1, the experimental set up:
To create a uniform magnetic field, the Helmholtz coils were positioned as shown above in Figure 1 and their electrical currents were tested to be equal in addition to flowing in the same direction.
The Hall probe was then used to measure the magnetic...

...Definition of Force
A force is a push or pull upon an object resulting from the object's interaction with another object. Whenever there is an interaction between two objects, there is a force upon each of the objects. When the interaction ceases, the two objects no longer experience the force. Forces onlyexist as a result of an interaction.
Velocity, Acceleration, Momentum, and Impulse
Velocity, in physics, is a vector quantity (it has both magnitude and direction), and is the time rate of change of position (of an object). However, quite often when you read ‘velocity’, what is meant is speed, the magnitude of the velocity vector (speed is a scalar quantity, it has only magnitude). For example: escape velocity (the minimum speed an object needs to escape from a planet, say); note that this can be easily turned into a velocity, by adding ‘in the direction radially out from the center of the planet’, and that this direction is sometimes implied (if not actually stated).
Velocity is a vector measurement of the rate and direction of motion or, in other terms, the rate and direction of the change in the position of an object. The scalar (absolute value) magnitude of the velocity vector is the speed of the motion. In calculus terms, velocity is the first derivative of position with respect to time.
The most common way to calculate the constant velocity of an object moving in a straight line is with...

...Composition of Concurrent Forces
Abstract— The experiment aims to show the relationship of different forces acting at the same time by using different variables namely, the equilibrant force, the resultant, and the given forces Fa and Fb. The equilibrant force is the force that establishes equilibrium while the resultant is the addition of the given forces and is also the opposite of the equilibrant. The objective of the experiment is to determine the resultant of two forces experimentally, to be checked by using its components and graphically adding the forces. The resultant obtained by component and graphically adding the vectors was approximately close to the resultant obtained experimentally. Though the results were close to each other, the different methods still resulted to slight differences. It is suggested that different approaches be used to give the most precise value.
Introduction
* The objective of the experiment is to determine the force which balances two other forces using the Force Table- this force is other wise known as the equilibrant force. In the experiment, the addition of the two forces was the resultant. The resultant should be equal to the equilibrant but opposite in direction. There are two methods of adding vectors, by graphically...

...MECHANICAL ADVANTAGE
CIRCUITS
SERIES CIRCUITS
Ratio of resistance and effort forces
Work = Force * Distance
IMA=
effort distance
resistance distance
AMA = resistance force
effort force
If MA > 1: Less effort force; greater effort
distance
If MA < 1: Greater effort force; less effort
distance
Moment = Force * Distance
Static Equilibrium: Effort Moment = Resistance Moment
Torque: A force that produces or tends to produce rotation or torsion.
SIMPLE MACHINES
Series circuits are connected end- Parallel circuits have both ends of
to-end with a single path for current the components connected together,
to flow.
with multiple paths for the current to
flow.
Current through every series
component is equal.
Voltage across every parallel
component is equal.
Total resistance is sum of
component resistances.
Total resistance is equal to the
reciprocal of the sum of the
Sum of all voltage drops is equal
component resistances' reciprocals.
to total applied voltage (KVL).
MULTIMETER USAGE
1
PARALLEL CIRCUITS
When measuring amperage, use
series connection. -->
CLASS 3
MA is always < 1
THERMODYNAMICS
The study of the effects of work, heat flow, and energy on a system.
Thermal energy: Kinetic energy in transit from one object to another due to
temperature differences.
Temperature: the average kinetic energy of...

...Force vs. Area
Connor Blackmon
Chemistry I H, 1st Period
Mrs. Kris Clements
October 18, 2012
Problem
Will a balloon pop if it is places on a bed of nails and pressure is applied?
Hypothesis
If a balloon is placed on a bed of nails and a force is applied, then the balloon will burst.
Variables
Independent variable- Force applied to the balloon and number of nails
Dependent variable- Does the balloon burst?
Materials
14 inch by 14 in by .75 in plywood board x2
196 nails
4 rods (14 inches tall)
Ruler
Pen
Drill
10 latex and 10 rubber balloons
Weights (1 lb, 5 lbs, 10 lbs; multiple of each weight)
Procedures:
Assembling the Board:
Using a pen and ruler, every one inch make a mark on one of the boards, these marks should be parallel to each other
use a drill to place a nail at each one of the points made on the board, all nails will be used
On the four corners drill a hole for a 14 inch rod facing the same way as the nails
Using the drill again, make four holes in the corners of the other plywood board for the rods to slide through
Experiment Procedures:
Inflate the rubber balloons to 11 inches in diameter, all balloons should be plus or minus .2 of an inch in diameter
Place rubber balloon on the middle of the bed of nails
Slide plywood board trough the rods to sandwich to balloon
Record if the balloon pops or not and weight applied to balloon...

...1. Which of the following sets of horizontal forces could leave an object in equilibrium?
a. 25, 50 and 100 N
b. 5, 10, 20 and 50 N
c. 8, 16, and 32 N
d. 20, 20 and 20 N
2. Which of the following sets of horizontal forces could not leave an object in equilibrium?
a. 6, 8 and 10 lb
b. 10, 10 , and 10 lb
c. 10, 20 and 30 lb
d. 20, 40 and 80 lb
3. If an object is free to move in a plane, the number of scalar equations that must be satisfied for it to be in equilibrium is
a. 2
b. 3
c. 4
d. 6
4. If an object is free to move in three dimensions, the number of scalar equations that must be satisfied for it to be in equilibrium is
a. 2
b. 3
c. 4
d. 6
5. An object in equilibrium may not have
a. Any forces acting on it
b. Any torques acting on it
c. Velocity
d. Acceleration
6. Two ropes are used to support a stationary weight W. The tensions in the ropes must
a. Each be W/2
b. Each be W
c. Have a vector sum of magnitude W
d. Have a vector sum of magnitude greater that W
7. A weight is suspended from the middle of a rope whose ends are at the same level. In order for the ends of the rope to be perfectly horizontal, the forces applied to the ends of the rope
a. Must be equal to the weight
b. Must be greater than the weight
c. Might be so great as to break the rope
d. Must be...

...Chapter 1 Vectors, Forces, and Equilibrium
1.1 Purpose
The purpose of this experiment is to give you a qualitative and quantitative feel for vectors and forces in equilibrium.
1.2
Introduction
An object that is not accelerating falls into one of three categories: • The object is static and is subjected to a number of diﬀerent forces which cancel each other out. • The object is static and is not being subjected to anyforces. (This is unlikely since all objects are subject to the force of gravity of other objects.) • The object is moving with constant velocity. In this case, the object may be subject to a number of forces which cancel out or no force at all. This case is not considered in this lab. The category of physics problems that involve forces in static equilibrium is called statics. Physicists and engineers are subjected to static problems quite frequently. A few examples of these principles in use are seen in the design of bridges and the terminal velocity of a person falling through the air. Mathematically, forces in equilibrium are just a special case of Newton’s Second Law of Motion, which states that the sum of all forces is equal to the mass of the object multiplied by the acceleration of the object. The special case of forces in equilibrium (static), occurs when the acceleration of the object is...

...Force in effect when car brakes
A car of mass m=1200 kg is traveling at a speed of 50km/h. Suddenly the brakes are applied and the car is brought to a stop over a distance of 20m. Assuming constant breaking force find:
(1) the magnitude of the breaking force,
(2) the time required to stop.
(3) What will be the stopping distance if the initial speed is 100km/h?
Solution.
Most of problems from Dynamics can be seen as “two parts problem”, one involving kinematics and the other - dynamics. This is a consequence of Newton’s Second Law - Force is a product of mass and acceleration.
Acceleration by itself is a purely “kinematical” problem. When mass is involved, we go into Dynamics.
In our problem the following are given:
m = 1200 kg – mass of the car,
v1 = 50 km/h – initial speed in the first case,
D1 = 20m – stopping distance in the first case,
v2 = 100 km/h – initial speed in the second case.
We are suppose to find:
F = ? – magnitude of breaking force,
t = ? – the time required to stop,
D2 = ?<="" p="">
We write down formulas which involved the unknown quantities,
F = ma (1)
a = v1/t (2)
D1 = v1t –(1/2) a t2 (3)
Some explanations:
Formula (1) is simply Newton’s Second Law of Motion,
formula (2) – the speed decreases from v1 to 0 during time t. Assuming constant breaking force means constant acceleration (deceleration or acceleration...