Magnetics
11.0 Magnetic force on a current carrying conductor
11.0 Magnetic force on a current carrying conductor
When a current-carrying conductor is placed in a magnetic field, magnetic force is acted upon the moving charges $($electrons$)$ within the conductor. This force is transmitted to the material of the conductor, and the conductor as a whole experience a force distributed along its length. Suppose a conducting wire of length $l$ carrying current $i,$ lies in a magnetic field $\vec B$ see the fig.
Consider a small element of length $\overrightarrow {dl} $ of the wire. The free electrons drift with a speed $v_d$ opposite to the direction of the current. We know that $i = jA = neA{v_d}$ Here $A$ is the area of cross-section of the wire and $n$ is the number of free electrons per unit volume. Each electron experiences an average magnetic force$${f_m} = - e\left( {{{\vec V}_d} \times \overrightarrow B } \right)$$The number of free electrons in the element is $nAdl.$ Thus the magnetic force on the wire is$$d{{\vec F}_m} = \left( {nAdl} \right){{\vec f}_m}$$$$ = \left( {nAdl} \right)\left[ { - e\left( {{{\vec V}_d} \times \overrightarrow B } \right)} \right]$$
If we take $\vec dl$ along the direction of the current, then ${\vec v_d}\left( {dl} \right) = {v_d}\left( {d\vec l} \right)$ and the above expression becomes$$d{{\vec F}_m} = \left( {nAe{v_d}\left( {d\vec l \times \vec B} \right)} \right)$$$$d{{\vec F}_m} = i\left( {d\vec l \times \vec B} \right)$$$$d{{\vec F}_m} = id\vec l \times \vec B$$
Quantity $id\vec l$ is called current element. The force on the wire of length $l$$${\vec F_m} = i\vec l \times \vec B$$
The above equation can be used in two stages
To find the magnitude of force by $F = B\;il\sin \theta $ and the direction of force can be obtained by Fleming left-hand rule or by right-hand palm rule.
Force on curved conductor
Let us consider a conducting wire of arbitrary shape and is placed in uniform magnitude field $\vec B$ The force on $d\vec l$ length of the conductor $d{\vec F_m} = \left( {nAdl} \right){\vec f_m}$ To get force on the whole wire, integrate above equation over the length of the wire. Thus$${{\vec F}_m} = \int_P^Q {id\vec l \times \vec B} $$$${{\vec F}_m} = i\left[ {\int_P^Q {d\vec l} } \right] \times \vec B$$$${{\vec F}_m} = i\overrightarrow {PQ} \times \vec B$$The other simpler way to get the force on current carrying wire is : draw straight line joining the ends of the conductor $($here $PQ),$ and then find its component perpendicular $\vec B$ here it is $PQ \sinθ.$
Therefore$$F = Bi\left( {PQ\sin \theta } \right).$$
