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:

Magnetic force on conductor: \[ \vec{F} = I (\vec{L} \times \vec{B}) \] Direction by Fleming’s left-hand rule.

Answer 1

Concept: A current-carrying conductor in a magnetic field experiences a magnetic force due to the motion of charge carriers. Force on a moving charge: \[ \vec{f} = q (\vec{v} \times \vec{B}) \]
Step 1: Consider free electrons. Let:

Number density of electrons = \( n \)
Drift velocity = \( \vec{v}_d \)
Volume of conductor = \( AL \)
Total number of electrons: \[ N = nAL \]
Step 2: Force on one electron. \[ \vec{f} = -e (\vec{v}_d \times \vec{B}) \] (Minus sign for electron charge)
Step 3: Total force on all electrons. \[ \vec{F} = N \vec{f} = nAL (-e)(\vec{v}_d \times \vec{B}) \] \[ \vec{F} = -neAL (\vec{v}_d \times \vec{B}) \]
Step 4: Use current relation. Current: \[ I = neA v_d \] So: \[ neA \vec{v}_d = I \hat{l} \] (where \( \hat{l} \) is direction of current) Substitute: \[ \vec{F} = -L (I \hat{l} \times \vec{B}) \]
Step 5: Direction convention. Force direction is defined using conventional current direction (opposite electron motion). Thus: \[ \vec{F} = I (\vec{L} \times \vec{B}) \] Where \( \vec{L} \) is a vector along the conductor of magnitude \( L \). Final Expression: \[ \boxed{\vec{F} = I (\vec{L} \times \vec{B})} \] Magnitude: \[ F = ILB \sin \theta \] where \( \theta \) is the angle between current and magnetic field. Conclusion: A current-carrying conductor experiences a magnetic force perpendicular to both current direction and magnetic field.