Question

Derive an expression for the force \(\vec F\) acting on a conductor of length \(L\) and area of cross-section \(A\) carrying current \(I\) and placed in a magnetic field \(\vec B\).

Hint:

For a wire of any shape placed in a uniform magnetic field, \[ \vec F=I(\vec L\times\vec B), \] where \(\vec L\) is the vector joining the initial and final points of the wire.

Answer 1

Force on a Current Carrying Conductor

Step 1: Consider a conductor placed in a magnetic field. Let \[ n \] be the number density of free electrons. In a small volume \[ AL, \] number of electrons is \[ N=nAL. \]

Step 2: Magnetic force on one electron. Lorentz force is \[ \vec F_e=-e(\vec v_d\times \vec B). \] Magnitude: \[ F_e=e v_d B\sin\theta. \]

Step 3: Force on all electrons. \[ F=N e v_d B\sin\theta. \] Substituting \(N=nAL\), \[ F=nALev_dB\sin\theta. \] Using \[ I=neAv_d, \] we obtain \[ F=BIL\sin\theta. \] Hence \[ \boxed{ F=BIL\sin\theta } \] and in vector form \[ \boxed{ \vec F=I(\vec L\times\vec B) } \]