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

FL =B I ℓ            (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
  • ℓ      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 1

B is parallel to  I :  magnetic force  FL = 0 N       

Situation 2

Situation 2

           

FL = BIℓ           

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
  • ℓ = 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’ℓ = B  I x sin α  ℓ = (0.20)( 4.0 x 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.    

Enable Notifications    OK No thanks