Force on a Current-Carrying Conductor Placed in a Uniform Magnetic Field

Electric motors, loudspeakers, and galvanometers all work because a current-carrying wire placed in a magnetic field experiences a push. Understanding this force explains how electrical energy becomes mechanical motion in everyday devices.

Materials: flexible copper wire, stand, electric cell, horseshoe magnet.

Steps:

1. Fix the copper wire on a stand so it passes between the poles of the horseshoe magnet.
2. Connect the wire to the cell to complete the circuit.
3. Observe the wire with no current flowing — it stays straight (position A).
4. Switch on the current flowing from top to bottom — the wire bends to position C.
5. Reverse the current direction — the wire bends to position B instead.

Observation: A current-carrying wire placed in a magnetic field experiences a force. Reversing the current reverses the force direction. This confirms the force depends on both the current's direction and the magnetic field's direction.

A current is simply electrons drifting through a conductor, though by convention we describe "current direction" as opposite to electron flow. When this moving charge sits inside a magnetic field, each electron feels a small magnetic push; adding up these microscopic pushes gives one net macroscopic force on the whole wire.

Both rules describe the same physical result — force perpendicular to both current and field — but use different hands and finger assignments, so always double-check which rule a question or teacher expects.

1. Consider a wire segment of length l and cross-sectional area A, carrying current I, placed in field \[\vec{B}\].
2. Each free electron drifting at velocity v_d​ feels a force F′ = ev_dB sin ⁡θ, where θ is the angle between the wire and \[\vec{B}\].
3. The number of free electrons in this segment is N = nAl, where n is the electron density.
4. Total force on the segment is F = F′ × N = (nev_dA)Bl sin ⁡θ.
5. Since nev_dA = I (the current), this simplifies to: F = BIl sin⁡ θ   ...(i)
   In vector form: \[\vec F\] = I\[\vec l\] × \[\vec B\]   ...(ii)

• When θ = 0 (wire parallel to field): sin⁡θ = 0, so the force is zero. This defines the direction of the magnetic field itself — the direction along which a current feels no force.
• When θ = 90 (wire perpendicular to field): sin⁡θ = 1, so force is maximum, F_max = BIl.
• For a wire of arbitrary shape from point P to Q, only the straight displacement vector matters: \[\vec F\] = I(\[\overrightarrow {PQ}\] × \[\vec B\]).

A horizontal rod of length 0.45 m and mass 60 g is suspended by two vertical wires. A current of 5.0 A flows through it. (g = 9.8 m/s²)

a. Find the magnetic field needed for zero tension.

For zero tension, the magnetic force must balance the weight exactly:

b. Find total tension if the current direction is reversed.

Reversing the current reverses the magnetic force direction, so now both weight and magnetic force pull downward:

• A current-carrying conductor in a magnetic field experiences a force perpendicular to both the current and the field direction.
• Reversing current or reversing field polarity reverses the force direction.
• Formula: F = BIl sin ⁡θ; vector form \[\vec F\] = I\[\vec l\] × \[\vec B\].
• Force is zero when the wire is parallel to \[\vec{B}\] and maximum (F_max = BIl) when perpendicular.
• Fleming's Left-Hand Rule (thumb = force, forefinger = field, middle finger = current) is the standard tool for direction in Indian board exams.