Question

Derive the expression for the force on a current-carrying conductor in a magnetic field.

Answer 1

1. The force experienced is equal to the sum of Lorentz forces on the individual charge carriers in the wire.
2. Consider a small length of wire ‘dl’, area – A, Current – I.
3. The free electrons drift opposite to the direction of the current.
4. The relation between I & V_d is given by
   I = neAV_d 
   
5. In magnetic field, the force experienced is given by
   `vec"F" = - "e" (vec"v"_"d" xx vec"B")`
   n → number of free electrons per unit volume.
   n = `"N"/"V"`
   V = A dl
6. Lorentz force = n A dl × – `"e"(vec"v"_"d" xx vec"B")`
   dF = -enA dl `(vec"v"_"d" xx vec"B")`
7. Current element, `"I"vec"dl" = - "enA"vec"v"_"d"`
   `therefore "d"vec"F" = ("l" vec"dl" = vec"B")`
   `vec"F" = ("l"  vec"l" xx vec"B")`
   F = BIl sin θ
   Case (i):
   If the conductor is along the magnetic field; θ = 0° ; F = 0.
   Case (ii):
   If the conductor is perpendicular to the magnetic field; θ = 90°; F = BIl.