·        Report to 1^st per. class; then, ESE and ESOL students report to their extended

         time testing locations for the second session of the Reading FCAT; and grade 11

         students report to testing locations for the FCAT Science test.

         changes have been made) report to former testing location from 3/10 and 3/11

         for an extended time. 

1. The force F produced by a magnetic field on a single charge depends upon

the speed v of the charge, the strength B of the field, and the magnitude of

the charge q, with F = qvBsinθ. θ is the smaller angle between v and B. 

2. In vector form, this equation is given by the cross product F = qvxB . To

find the direction of the force, use the right hand rule.

3. The unit of magnetic field strength is the Newton per Ampere∙meter, N/Am,

or Tesla (T), named after Nicola Tesla (1856-1940), physicist/engineer/inventor

from Croatia. The non-metric unit is the Gauss (G), with 1 T = 10⁴ G.

4. If the charged particle moves parallel to the field lines (θ = 0), then the

magnetic force on the particle is zero. If a charged particle is moving

perpendicular to a uniform magnetic field, the path of the charged particle is

an arc (or circle).

5. The magnetic force is the source of the centripetal force on the charged

particle. This relationship can be used to find the radius of the arc when we

set the equations equal to one another.    mv²/r = qvB

6. Since the magnetic force is perpendicular to the velocity of the charged

particle, the force does not cause the speed of the particle to change, only

its direction. Thus, no work is done by the magnetic force on the charged

particle.

7. In regards to forces due to magnetic fields, Ampere found that a force is

exerted on a current-carrying wire in a magnetic field, F = BILsin θ, where

B is the magnetic field in Teslas (T), I is the current, L is the length of wire

in meters, and θ is the angle.

8. Only the perpendicular component of B exerts a force on the wire.

9. If the direction of the current is perpendicular to the field (θ = 90), then

the force is given by F = BIL.

10. Recall that B can be calculated by using the equation, B = F / I·L.

11. If a long straight conductor of length L carries a current I, the force on

that conductor when placed in a uniform magnetic field is calculated using

the cross-product, F = ILxB.

12. The net magnetic force on any closed loop carrying a current in a uniform

magnetic field is zero.

13. The magnetic moment of a current loop carrying a current I is, μ = IA,

where A is perpendicular to the plane of the loop and has magnitude equal

to the area.

14. The torque τ on a current loop when the loop is placed in an external

magnetic field is given by the equation, τ = μ x B.

15. When a current-carrying conductor is placed in a magnetic field, a potential

difference is generated in a direction perpendicular to both the magnetic field

and the current.

16. This potential difference, known as a transverse voltage, was first noticed

by Edwin Hall (1855-1938) in 1879 and is known as the Hall Effect.

17. Using the equations for Magnetic Force, Electric Field, and drift velocity

derived previously, we can show that the Hall Voltage measured across a

conductor of width d and cross-sectional area A is, ΔV_H = IBd/nqA.

1. Approximately, two out of every three years (according to what I have been

able to observe) one of the free response questions involves a situation where

a charge is accelerated through two charged plates.

2. The charge then enters a magnetic field whose direction causes the charge to

move in a circle.

3. Make sure you are able to do the following:

    (a) Calculate the speed of the charge as it exits the region between the two

        charged plates.

    (b) Draw the direction of the electric field between the two charged plates.

    (c) If there is also a magnetic field between the two charged plates in addition

        to the electric field, explain the relationship between the two fields that

        allows the charge to pass through undeflected.

    (d) Remember these three formulas for regions where both electric and

        magnetic fields exist: V=Ed, qE=F, and F=qvB.

    (e) Manipulating these formulas allows you to express the velocity of the

        charge in terms of E and B. This also allows you to write an expression for

        the accelerating voltage in terms of v, B, and d.

    (f) Make sure you can prove that this would also yield v = E / B .

    (g) Remember, if the charge is moving in a circle and the magnetic field is

        perpendicular to it, it does no work on the charge. It only changes its

        direction.

    (h) Remember, if the charge is moving in a circle, the magnetic force provides

        the centripetal force. This allows you to calculate the radius.

    (i) Sometimes, you are asked to calculate the thermal energy dissipated by

        the accelerated charge if it is allowed to strike a target. Since you know its

        speed, just calculate its energy.

    (k) Using a mass spectrometer, and knowing that the radius of the path of a

        particle is proportional to its mass, with several particles you can set up a

        proportion between their masses and radii to determine the mass of an

        unknown particle.

    (l) If the charged particle moves in a circular path, the centripetal force equals

        the magnetic force. This equality can be solved for the ratio of charge to

        mass of the particle (q/m).

    (m) More points are awarded if you express the direction in terms of positive

        or negative x, y, and z, using the unit vectors i, j, and k.

And still, we need these steps to solve any problem in Physics:

(i) read the problem and identify the given variables

(ii) determine what you are asked to solve for

(iii) find the correct equation to use

(iv) use Algebra, Trigonometry, and/or Calculus to isolate the unknown

(v) substitute-in the given information and simplify.