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
   ${\displaystyle \overrightarrow{\text{F}}=-\text{e}({\overrightarrow{\text{v}}}_{\text{d}}\times \overrightarrow{\text{B}})}$
   n → number of free electrons per unit volume.
   n = ${\displaystyle \frac{\text{N}}{\text{V}}}$
   V = A dl
6. Lorentz force = n A dl × – ${\displaystyle \text{e}({\overrightarrow{\text{v}}}_{\text{d}}\times \overrightarrow{\text{B}})}$
   dF = -enA dl ${\displaystyle ({\overrightarrow{\text{v}}}_{\text{d}}\times \overrightarrow{\text{B}})}$
7. Current element, ${\displaystyle \text{I}\overrightarrow{\text{dl}}=-\text{enA}{\overrightarrow{\text{v}}}_{\text{d}}}$
   ${\displaystyle \therefore \text{d}\overrightarrow{\text{F}}=(\text{l}\overrightarrow{\text{dl}}=\overrightarrow{\text{B}})}$
   ${\displaystyle \overrightarrow{\text{F}}=(\text{l} \overrightarrow{\text{l}}\times \overrightarrow{\text{B}})}$
   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.