Motional emf in a rod moving in a magnetic field
A rod of length l moving at speed v across a field B has an emf ε = Blv. The Lorentz force on its free charges causes it.
The set-up
Take a U-shaped conductor: two parallel rails joined at one end. It lies in a plane, and a uniform magnetic field B acts perpendicular to that plane. A straight rod PQ of length l rests across the rails and can slide on them. The rod and the rails form a closed rectangle. When the rod moves with velocity v, the area of the rectangle changes, so the flux changes. The emf produced in a conductor moving in a magnetic field is called motional emf. Fleming's right-hand rule gives the direction of the induced current: thumb along the motion, first finger along the field, middle finger along the current.
Derivation 1: from the change in flux
- 1Let x be the distance of the rod PQ from the closed end of the rails. The area of the loop is lx.
- 2Flux through the loop: Φ = Blx, since B is perpendicular to the plane.
- 3Induced emf: ε = −dΦ/dt = −Bl dx/dt.
- 4The rod moves towards the closed end with speed v, so x decreases: dx/dt = −v.
- 5★So, ε = Blv. This is the motional emf.
★ key step
Derivation 2: from the Lorentz force
- 1Each free charge q in the rod moves with the rod at velocity v, perpendicular to B.
- 2The Lorentz force on it has magnitude F = qvB and acts along the rod.
- 3Work done by this force in moving the charge from one end of the rod to the other: W = Fl = qvBl.
- 4emf is the work done per unit charge: ε = W/q.
- 5★So, ε = Blv, the same result. The magnetic force on moving charges is the cause of motional emf.
★ key step
Worked example: rod on rails
A rod 50 cm long slides at 4 m/s on rails in a uniform field of 0.2 T perpendicular to the plane of the rails. The circuit has a total resistance of 2 Ω. Find (a) the emf, (b) the current and (c) the current if the rod slides at 8 m/s instead.
- Given: l = 50 cm = 0.5 m, v = 4 m/s, B = 0.2 T, R = 2 Ω.
- (a) ε = Blv = 0.2 T × 0.5 m × 4 m/s = 0.4 V.
- (b) I = ε/R = 0.4 V ÷ 2 Ω = 0.2 A.
- (c) At 8 m/s: ε = 0.2 T × 0.5 m × 8 m/s = 0.8 V, so I = 0.8 V ÷ 2 Ω = 0.4 A. Doubling v doubles the emf and the current.
- By Lenz's law, the induced current in the rod feels a magnetic force that opposes the motion, so a steady push is needed to keep v constant.
(a) 0.4 V; (b) 0.2 A; (c) 0.4 A
Harder example: a rod rotating about one end
A metal rod 0.5 m long rotates at 200 rad/s about an axis through one end. The axis is parallel to a uniform field of 0.4 T, so the rod sweeps a plane perpendicular to the field. Find the emf between the ends of the rod.
- Different parts of the rod move at different speeds: zero at the fixed end and ωl at the free end.
- Average speed of the rod = (0 + ωl)/2 = ωl/2.
- ε = Bl × (average speed) = Bl × ωl/2 = ½Bωl².
- ε = ½ × 0.4 T × 200 rad/s × (0.5 m)² = 0.2 × 200 × 0.25.
- ε = 40 × 0.25 = 10 V.
emf = 10 V
How to draw the set-up
- Draw two long parallel horizontal lines for the rails, joined at the left end by a short vertical line with a resistor R in it.
- Draw the rod PQ as a vertical line across the rails and label its length l.
- Fill the region with crosses (×) for B into the page, or dots (•) for B out of the page, and label B.
- Draw an arrow on the rod showing its velocity v.
- Mark x, the distance from the closed end to the rod.
- Show the direction of the induced current round the loop with arrows, found by Fleming's right-hand rule.
When an aircraft flies over India, its metal wings cut the vertical component of Earth's magnetic field. A small motional emf, ε = Blv, appears between the wing tips.
3 marks: “Using the Lorentz force, derive an expression for the emf induced across a rod of length l moving with velocity v perpendicular to a uniform field B.” Force on each free charge qvB along the rod; work over length l is qvBl; emf = work per unit charge = Blv.
2 marks: “A 1.5 m long rod moves at 10 m/s at right angles to a field of 0.04 T. Find the emf. What is the emf if it moves parallel to the field?” ε = Blv = 0.04 T × 1.5 m × 10 m/s = 0.6 V; moving parallel to the field, ε = 0.
5 marks: “(a) A rod of length l slides with velocity v on rails in a uniform field B perpendicular to their plane. Derive the motional emf from the change in flux. (b) A metal rod 1 m long rotates at 50 rad/s about one end, in a plane perpendicular to a uniform field of 0.2 T. Find the emf between its ends.” (a) Φ = Blx; ε = −dΦ/dt = −Bl dx/dt = Blv, since dx/dt = −v. (b) ε = ½Bωl² = ½ × 0.2 T × 50 rad/s × (1 m)² = 5 V.
Motional emf is Blv: field × length × velocity, when all three are at right angles.