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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

 

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

electric_field

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

 

magnetic_force_2

magnetic_force_1

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

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

magnetic_force_3

          

           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

     

Left_Hand_Rule

        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

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_of_a_current_carrying_wire

        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