7. Force on a current carrying wire in uniform magnetic field

How much force a uniform magnetic field applies on a current carrying conductor

Magnetic force on a current-carrying conductor

Let a current \( I \) flow in a conductor having cross-sectional area \( A \). The number density of mobile charge carriers (electrons) is \( n \) and their drift velocity is \(\vec {v_d} \). The conductor is placed in a magnetic field zone \( \vec{B} \).

Consider a small current element \( I \, \vec{dl} \) in the conductor making an angle \( \theta \) with \( \vec{B} \). The charges present in the current experience a force \( F_B = q (\vec{v_d} \times \vec{B}) \).

Total volume of the current element of length \( dl \) is \( A \, dl \). Number of charge carriers present in the given volume is \( n \, A \, dl \).

Total force on the current element

The total force experienced by the current element \( I \, d\vec{l} \) is the sum of individual forces on all these charges. Net force on the current element:
\[ d\vec{F_B} = n \, A \, dl \, q (\vec{v_d} \times \vec{B}) \]

Since \( nq \vec{v_d}= \vec{J} \) (current density):
\[ d\vec{F_B} =\vec {J} \, A \, dl \times \vec{B} \]
Also:
\[\vec{ J} \, A dl= Id\vec{l} \]
Hence:
\[ d\vec{F_B} = I \, d\vec{l} \times \vec{B} \]

Force on a straight current-carrying conductor placed in a uniform magnetic field

For a straight conductor of length \( l \) carrying current \( I \) in a magnetic field \( \vec{B} \):
\[ \vec{F} = I \, \vec{l} \times \vec{B} \]

If the conductor makes an angle \( \theta \) with \( \vec{B} \):
\[ F = I \, l \, B \, \sin \theta \]