FE Physics
Formulas & Explanation     Electric Field

E = Fel/q

E     electric field in N/C

Fel  electric force in N

q     electric charge in C

Example

A charge of  2.0 x 10-6 C is placed at a certain place in an electric field

The electric force on that charge is 1.0 x 10-3 N.

Calculate the magnitude of the electric field at that place.

E = Fel/q

E = (1.0 x 10-3)/(2.0 x 10-6) = 5.0 x 102 N/C

In a uniform electric field is the field of the same magnitude and direction at all places.

Acceleration of a charged particle in a uniform electric field.

Two formulae are important.

E = Fel/q

(significance of the symbols: see above)

Δ Ekin  = q ΔV Δ Ekin  change of the kinetic energy in  J

q          charge in C

ΔV       potential difference in  V or J/C

Example

Electrons are accelerated between the negative plate K

and the positive plate  A

The potential difference between the plates is  2.0 x 102 V

The electric field is 5.0 x 102 N/C

The velocity  speed of the electrons at K is negligible.

a.       Calculate the electric force on the electrons

b.      Find the speed of the electrons at A.

a.   E = Fel /q

Fel = q E = (1.6 x 10-19)(5.0 x 10 2)= 8.0 x 10-17 N.

b.   q.ΔV = Δ Ekin

(At acceleration  : q.ΔV > 0 ;   at delay :  q.ΔV <0)

(1.6  X 10 -19)( 2.0 x 102 ) = Ekin,A – Ekin,K

3.2 x 10-17 = ½ mvA2 – 0

3.2 x 10-17 = (½) (9.1 x 10-31) vA2

vA 2 = (3.2 x 10-17)/4.55 x 10-31

vA  = 8.4 x 106 m/s

Magnetic force on a current-carrying wire in a magnetic field

FL = B I             (B I )

FL    magnetic force (or Lorentz force) in N

B     magnetic induction in T (Tesla)           1 T = 1 N.A-1 .m-1

I      current in A

length of the wire  in the magnetic field

Example 1

A long straight wire is in a uniform magnetic field

The electric current is 3.0  A

B = 5.0 x 10-2 T

What is the magnitude of the magnetic force in the two situations below and what direction has this force?

Situation 1                                                                                    Situation 2  B is parallel to  I :  magnetic force  FL = 0 N                   FL = BI

FL  = (5.0 x 10-2 )(.3.0)(0.40) = 0.06 N

Find direction with for example using the Left-Hand-Rule

-catch the field lines on the palm of your left

hand

- fingers in the direction of I

- thumb  gives the direction magnetic force F

FL  is perpendicular to B and I

In this example  : F  is perpendicular to the plane of the paper and points to you.

This is indicated by a dot.

Example 2

Find the magnetic force in the next situation B = 0.20 T

I = 4.0 A

= 25 cm  (length wire  in  magnetic field)

α = 40 o

Resolve the vector I into two components  .

One component parallel to the vector B and the other

component (I’) perpendicular to B

sin α = I’/I

I ‘= I x sin α

The formula to calculate the magnetic force :

FL = B I’ = B * I x sin α * = (0.20 )( 4.0) ( sin 40o )(0.25) = 0.13 N

Remark : One can also resolve B into two components parallel and perpendicular to I.

Then use the component perpendicular to I.

The motion of a charged particle in a uniform magnetic field

FL  = B q v

FL       magnetic force  in N

B      magnetic induction in T

q      charge particle in C

v     speed particle in m.s-1

Example   1

An electron enters at A in a uniform magnetic field with a speed of  4.0 x 107  m.s-1

The magnetic field has a magnitude of  3.0 x 10-3 T  and the direction is into the paper

The electrons move perpendicular to the magnetic field.

On the electron acts a magnetic force :

FL   = B q  v FL   = (3.0 x 10-3 )( 1.6 x 10-19 )( 4.0 x 107 )= 1.92 x 10-14 N

The Left-Hand-Rule determines the

direction of the magnetic force . (For the description of

this rule : see example 1 in the previous

paragraph)

The electron goes at A from the left side to the right

side.

The direction of the current I is opposite to the

direction   of  the velocity of the electron

So the direction of I is from right to left

The direction of the magnetic force is downward.

F  is perpendicular to the velocity v

The electron is in a uniform circular motion.

Example 2

Calculate the radius of the path

FL is the centripetal force

FL  = Fc

B q v = (m v2)/r

r = (m v)/(B q)

r = (9.1 x 10-31 *  4.0 x 107) / (3.0 x 10-3 * 1.6 x 10-19 ) = 7.6 x 10-2 m

Magnetic field of a long current-carrying coil B = μo . N/. I

B    magnetic induction in T

μo    magnetic permeability in vacuum (air)

μo = 4π x 10-7 Hm-1

N     number of turns

length coil in m

I       electric current in  A

Example

A coil contains 200 turns and has a length of 40 cm.

The electric current is 0.30 A

Find the magnetic induction in the coil

B = μo ( N/ℓ) I

B =  (4π x 10-7 )  (200/0.40) (0.30) = 9.4 x 10-5 T

We find the direction of the magnetic induction with the Right-Hand-Rule

-  fingers of the right hand in the direction of I

-  the thumb points in the direction of the magnetic  induction B ( see the figure above)

Remark:

IN the coil the field lines are directed from  Z to N

So at the left side of the coil is the magnetic N pole and at the right side the magnetic S pole

Magnetic field of a current-carrying wire

Right-Hand-Rule

Point the thumb of the right hand in the direction of the current

The tips of curled fingers will point in the direction of the magnetic field B.

B =( μo I)/ (2πr)

r     perpendicular distance to the wire in m

Example

A long straight wire carries a current of 16 A.

The magnetic induction in a point is 4.0 x 10-5 T

Find the (perpendicular) distance  to the wire

B =( μo I)/ (2πr)

4.0 x 10-5 = (4π x 10-7  x 16)/(2πr)

2πr * 4.0 x 10-5 = 64π x 10-7

r = (64 x 10-7)/(8.0 x 10-5) =  0.08 m