FE Physics
 Formulas & Explanation
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Electric Field


E = Fel/q


E     electric field in N/C

Fel  electric force in N

q     electric charge in C



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



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                   


- 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                       


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


          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

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



         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)



        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    


        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 




        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