Force on current carrying conductor in uniform magnetic field

Derivation of force on current carrying conductor in uniform magnetic field:

Let us consider:
  • The length of the conductor - $l$

  • The cross-section area of the current carrying conductor - $A$

  • The current flow in a conductor- $i$

  • The drift or average velocity of the free electrons - $v_{d}$

  • The current carrying conductor is placed in a magnetic field - $B$

  • The total number of free electrons in the crrent carrying conductor - $N$

  • Force on current carrying conductor in Uniform magnetic field
    Force on current carrying conductor in the uniform magnetic field

    Now the magnetic force on one free electron in a conductor -

    $F'= ev_{d}B sin\theta \qquad(1)$

    The net force on the conductor is due to all the free electrons present in the conductor

    $F=N\: F' \qquad(2)$

    Let $n$ is the number of free electrons per unit volume of conductor. So the total number of free electrons in the $Al$ volume of the conductor will be

    $N=nAl \qquad(3)$

    Now substitute the value of $N$ and $F'$ from above equation $(1)$ and equation $(3)$ in equation $(2)$

    $F=neAlv_{d}B\: sin\theta$

    Where $i=neAv_{d}$

    So from the above equation
    $F=ilB \: sin\theta $
    The Vector Form of the above equation:
    $F=i \left(l \times B \right) $