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A semicircular wire carrying current sits in a magnetic field, and the question asks for the force on it. Many students multiply by the arc length πR and lose the marks. In a uniform field there is a cleaner route: the bent wire feels exactly the same force as a straight wire joining its two ends. This post shows why, works three examples, and marks where the shortcut stops working.
What does the syllabus say?
The JEE (Main) 2026 syllabus lists "force on a current-carrying conductor in a uniform magnetic field" in Unit 13. The JEE (Advanced) 2026 brochure lists force on a current-carrying wire in a uniform magnetic field, and the CBSE Physics curriculum for 2026-27 has the same topic in Class XII Chapter 4. Check the syllabus for your exam year.
Why does a bent wire act like a straight one?
The NCERT chapter Moving Charges and Magnetism derives F = Il × B for a straight rod (Eq. 4.4), where l points along the current. For a wire of arbitrary shape, it says to treat the wire as a collection of small straight strips dl and add up I dl × B.
Now use the word uniform. If B is the same everywhere, it can come out of the sum:
F = Σ I dl × B = I (Σ dl) × B.
The small vectors dl, placed head to tail, add up to the straight vector from the start of the wire to its end. Call it L_AB. So
F = I L_AB × B, for any shape, in a uniform field.
Two results follow at once. A closed loop has L_AB = 0, so the net force on it is zero. And the force is always perpendicular to both B and the line joining the ends.
Worked example 1: a semicircle
A semicircular wire of radius 10 cm carries 5.0 A in a uniform field of 0.20 T perpendicular to its plane. Find the force.
The ends are a diameter apart, so L_AB = 2R = 0.20 m, perpendicular to B.
F = I(2R)B = 5.0 × 0.20 × 0.20 = 0.20 N, in the plane of the wire and perpendicular to the diameter.
Check by integration. A small piece at angle φ around the arc has length R dφ and feels IBR dφ, directed along the radius. The components along the diameter cancel in pairs. The components perpendicular to it add: ∫ IBR sin φ dφ from 0 to π = 2IBR. Same answer. The arc-length guess, IπRB ≈ 0.31 N, is too big.
Worked example 2: an L-shaped wire
A wire runs 3.0 cm along +x and then 4.0 cm along +y, carrying 2.0 A. The field is 0.50 T along +z.
Piece by piece:
First piece: x̂ × ẑ = −ŷ, so the force is 2.0 × 0.030 × 0.50 = 0.030 N along −y.
Second piece: ŷ × ẑ = +x̂, so the force is 2.0 × 0.040 × 0.50 = 0.040 N along +x.
Resultant: √(0.030² + 0.040²) = 0.050 N.
Shortcut: the ends are √(3² + 4²) = 5.0 cm apart, so F = 2.0 × 0.050 × 0.50 = 0.050 N, perpendicular to the line joining the ends. Both routes agree.
Worked example 3: when the net force is zero
Take the semicircle from example 1 but turn the field so it lies in the plane of the wire, parallel to the diameter. Now L_AB is parallel to B, so the net force is zero.
That does not mean nothing happens. Pieces on one half of the arc are pushed out of the plane and pieces on the other half are pushed in, so the wire feels a torque. The NCERT chapter treats this idea for a full loop in its section on torque on a current loop. Zero net force and zero torque are separate questions.
When does the shortcut fail?
Only in a non-uniform field. Near a long straight wire, for example, B changes with distance, so it cannot come out of the sum. Then you must integrate piece by piece. Our post on the Biot-Savart and Ampere laws shows how such fields vary.
Which mistakes cost marks?
Using arc length. Use the straight distance between the ends.
Getting the cross product backwards. Write Il × B, not B × Il. Our post on right-hand-rule mistakes has a method for directions.
Assuming zero force means no effect. Check torque separately.
Applying the shortcut in a non-uniform field.
Forgetting the angle. Only the component of L_AB perpendicular to B counts: F = IL_AB B sin θ.
A practice routine
Day 1: Derive F = I L_AB × B from the NCERT sum without notes.
Day 2: Redo the semicircle by integration and by the shortcut.
Day 3: Solve three shapes: a quarter circle (answer I√2 RB), a zigzag and a parabola-shaped wire.
Day 4: Mix in two torque-on-loop problems and say for each whether the force, the torque or both are zero.
Day 5: Do a timed mixed set. Dunlosky and colleagues (2013) found practice testing and spaced practice the most widely useful study techniques.
When a direction or a sign keeps coming out wrong, the Fermi Problem Library has worked problems with a step-by-step solver that explains the reasoning rather than printing an answer.
Draw any wiggly wire between two points tonight, mark the straight line between its ends, and write down the force in a field of your choice.
