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

Force on a Current-Carrying Conductor in a Magnetic Field
Consider a conductor of length L and cross-sectional area A carrying a current I. Let the conductor be placed in a magnetic field B. We aim to derive an expression for the force acting on the conductor due to the magnetic field.

Step 1: Force on a Single Charge
The force on a single charge q moving with velocity v in a magnetic field B is given by the Lorentz force:
F = q (v × B)
In a conductor, the charges are electrons with drift velocity v_d. If n is the number density of electrons, the number of electrons in the conductor of volume V = A L is:
N = n × A × L
The total force on all moving charges is:
F_total = N × q × (v_d × B)
F_total = (n A L) e (v_d × B)

Step 2: Express in Terms of Current
The current I in the conductor is related to drift velocity v_d by:
I = n e A |v_d|
So, n e A v_d = I (vector along the direction of current)
Substitute this into F_total:
F = I (L × B)

Step 3: Final Expression
The force acting on the conductor is:
F = I (L × B)
where:
- L is a vector along the length of the conductor in the direction of current
- B is the magnetic field vector
- × denotes the vector cross product, so the force is perpendicular to both L and B

Conclusion:
The magnitude of the force is given by:
|F| = I L B sin θ
where θ is the angle between the conductor and the magnetic field. The direction of the force is given by the right-hand rule.