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UY1: Magnetic Force On A Curved Conductor

Compute net magnetic force on a wire made of straight and curved segments in a uniform magnetic field.

  • University Physics Year 1
On this page

Learning objectives

  • Analyse magnetic forces, induction, inductance, and alternating-current systems with consistent signs.
Why this matters + quick links

This page gives the UY1 working model/result for Magnetic Force On A Curved Conductor. You reuse it when you build fields/potentials by symmetry or superposition, and when you connect fields to forces, energy, and circuits.

1) At a glance

Use the magnetic force element

d vector F = I d vector l × vector B

and sum vector contributions from each segment.

  • Modelling context: steady current and (usually) uniform vector B so the integral reduces to geometry plus cross-product directions.
  • Strategy: break the wire into pieces where symmetry or constant direction makes the integral easy.

Prerequisites: Magnetic Force On A Current Carrying Conductor
Next uses: Force & Torque On Current Loop In Magnetic Field

2) Setup

Assume:

  • Uniform magnetic field vector B points out of the page.
  • Wire carries steady current I.
  • Geometry has one straight segment of length L parallel to vector B, a semicircle of radius R in the page, and another straight segment contributing in + j hat.

Sign convention: + j hat is upward in the diagram plane.

Common traps (components + symmetry)
  • Don’t assume every part of a curved wire contributes the same direction force; integrate or use symmetry carefully.
  • For arcs, pair points symmetrically to show one component cancels.
  • Segments parallel to vector B contribute zero because d vector ℓ × vector B = 0.

3) Core derivation/explanation

Straight segments

For any segment parallel to vector B,

vector F = I vector L × vector B = 0

so that segment contributes no magnetic force.

The other straight segment (perpendicular to vector B) contributes magnitude

F_straight = BIL

directed + j hat for the stated current direction.

Semicircular segment

Magnitude element:

dF = B I dl

With dl = R dθ and vertical component dF_y = dF sin θ,

F_y = BIR∫₀^π sin θ dθ = 2BIR

Horizontal components cancel by symmetry.

So total force:

vector Fₜₒₜₐₗ = BI(L + 2R) j hat

Checks (sanity)

  • If B → 0 or I → 0, the total force must go to zero.
  • Dimensional check: BI(L + 2R) has units of newtons.

4) Worked example(s)

Let B = 0.50 T, I = 4.0 A, L = 0.30 m, R = 0.10 m.

F = BI(L + 2R) = 0.50 × 4.0 × (0.30 + 0.20) = 1.0 N

Direction: + j hat (upward).

5) Practice set (with hints + answers)

  1. If B doubles and all else stays fixed, what happens to total force? Hint: force is linear in B. Answer: it doubles.

  2. For fixed B and I, compare force for R = 0.05 m and R = 0.15 m (same L). Hint: use L + 2R. Answer: larger R gives larger force by factor (L + 0.30)/(L + 0.10).

  3. Why do horizontal components from the semicircle cancel? Hint: pair points at angles θ and π-θ. Answer: equal magnitude, opposite x-direction components.

6) Summary + next steps

For mixed straight-curved conductors, do vector addition segment by segment. Symmetry often removes one component, making the final result much cleaner.

Next: Force & Torque On Current Loop In Magnetic Field Previous: Magnetic Force On A Current Carrying Conductor Back To Electromagnetism (UY1)