Force on a Straight Current-Carrying Conductor
For a straight segment carrying steady current I in uniform B, F = I(L cross B), where L points along conventional current and has magnitude equal to segment length.
Why this shows up in the exam
Balancing rods on inclines · Calculating force on wire segments · Finding current direction from mechanical equilibrium
Learn the idea
A current-carrying segment in a magnetic field feels a sideways force given by current times the length cross field. Moving charges inside the wire feel Lorentz forces, and their combined push is transferred to the conductor. Only the length component across the field contributes.
🧠 Memory hook: Conventional current cross B gives the wire force.
Get this one clearly and it pays off every single time it shows up in the paper. 🎯
Formulas & facts to keep ready
- F = I(L cross B) — vector force on a straight current segment in uniform B
- |F| = I L B sin(theta) — force magnitude for angle theta between current and field
How to approach it
- 1Draw conventional current and B
- 2Find I L cross B
- 3Combine with all mechanical forces component by component
Common slip-ups that cost marks
- •Using electron-flow direction for L
- •Using the whole wire length instead of the part in the field
- •Ignoring gravity or support forces in equilibrium
🌟 That's the whole idea — you've got this. Try the practice set below; every question you attempt makes it stick a little harder.
Original chapter practice
Original questions for this chapter, not past-paper questions or an exact mapping to this individual concept.
A charge of 2 microC moves perpendicular to a 3 T magnetic field at 4 x 10^5 m/s. Find the magnetic force.
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