Force on a current carrying conductor in magnetic field

We know that a moving charge in a magnetic field experiences a force. Electric current in a conductor is due to the drifting of free electrons in a definite direction in the conductor. When such a current carrying conductor is placed in a uniform magnetic field, each free electron experiences a force. Since the free electrons are constrained in the conductor, the conductor itself experiences a force. Hence, a current carrying conductor placed in a uniform magnetic field experiences a force.

Let us consider a conductor of length $l$ and cross sectional area $A$ placed at an angle $\theta$ in a uniform magnetic field $B$.

Force on a current carrying conductor in magnetic field: F=BIlsinθ

Let $I$ be the current passing through the conductor, $\overrightarrow{v_d}$ be the drift velocity of free electrons, $n$ be the electron density (i.e. number of electrons per unit volume) and $e$ be the charge of an electron. Then, Force on each electron, \[\overrightarrow{F_e}=e(\overrightarrow{v_d}×\overrightarrow{B})\] \[F_e=ev_dB\sin\theta\] Since $n$ is the number of electrons per unit volume, the total electrons in the length $l$ of the conductor will be $n×\text{volume of conductor}=n×Al$.

Hence, the total force $F$ on the conductor is equal to the force acting on all the free electrons inside the conductor while moving in the magnetic field. \[\therefore F=\text{total number of electrons}×F_e\] \[=(nAl)F_e\] \[=(nAl)(ev_dB\sin\theta)\] \[F=(neAv_d)(lB\sin\theta)\text{ __(1)}\] We know that current through conductor is related with drift velocity by the relation, \[I=neAv_d\text{ __(2)}\] From equations $\text{(1)}$ and $\text{(2)}$, \[F=(I)(lB\sin\theta)\] \[\therefore F=BIl\sin\theta\] This is the required expression for a force experienced by a current carrying conductor placed in the magnetic field.

In vector form, \[\overrightarrow{F}=I(\overrightarrow{l}×\overrightarrow{B})\] The vector $\overrightarrow{l}$ is taken in the direction of current. The direction of this force $\overrightarrow{F}$ is perpendicular to the plane containing $\overrightarrow{l}$ and $\overrightarrow{B}$.

Special Cases

  1. If $\theta=0°$ or $180°$, $F=0$. Hence, if a current carrying conductor is placed parallel or antiparallel to the direction of field. Then, no force is experienced by the conductor.
  2. If $\theta=90°$, $F=BIl$ which is the maximum value of force experienced by the conductor.

The relation $F=BIl\sin\theta$ is valid only when magnetic field is uniform over the whole length of the conductor.

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