Physics · Electromagnetic Induction · NEET
In a straight rod moving with one constant velocity v, every point moves at the same speed, so ε = Bvl uses that single v. In a rotating rod, the speed changes along the rod: v = ωx, so the pivot (x = 0) is still and the tip (x = R) moves fastest at ωR. You cannot use one v — you integrate, or use the average speed which is (0 + ωR)/2 = ωR/2. That average is where the ½ comes from. So ε = B × (average speed) × R = B × (ωR/2) × R = ½BωR².
R is the length of the rotating rod, measured from the pivot (axis of rotation) to the free end. If the rod is a spoke of a wheel spinning about its centre, R is the radius of the wheel and the EMF is between the centre and the rim. Always measure from the fixed pivot, because the pivot is where speed is zero.
You put the average speed of the rod, not the tip speed. Average speed = ωR/2. Then ε = B × (ωR/2) × R = ½BωR². A common wrong step is using the tip speed ωR, which gives BωR² — double the correct answer. The rod's speed rises linearly from 0 to ωR, so its average is exactly half the maximum.
Yes. A disc is like infinitely many rods (radii) side by side. Each radius has EMF ½BωR² between centre and rim, and they are all in parallel, so the EMF between the centre and the rim of the disc is still ½BωR². This is the Faraday disc (homopolar generator). Adding more radii does not add EMF because they are in parallel, not series.
They describe different set-ups. The rotating rod ε = ½BωR² is a rod spinning in a plane, cutting field lines the whole time, giving a steady (DC) EMF. The AC generator has a closed coil rotating so its flux changes as cos(ωt), giving an alternating EMF that peaks at ε₀ = NBAω. Do not mix them: a single rod with a free end has no closed loop of changing flux, so you use the motional-EMF result ½BωR².
A cycle wheel of radius 0.5 m is rotated with constant angular velocity of 10 rad/s in a region of magnetic field of 0.1 T which is perpendicular to the plane of the wheel. The EMF generated between its centre and the rim is:
Try the real previous-year questions from this chapter — each with the answer and a full solution.
ε = ½BωR², where B is the magnetic field perpendicular to the plane of rotation, ω is the angular speed in rad/s, and R is the length of the rod from the pivot to the free end. The EMF is measured between the pivot end and the free end.
Use the force on positive charges F = qv × B. The magnetic force pushes positive charge toward one end, making it the higher-potential end. The direction depends on the sense of B and the direction of rotation; reverse either one and the high-potential end flips.
A single rod spinning at constant ω in a steady field B produces a constant (DC) EMF of ½BωR², because it keeps cutting field lines at the same rate. This is unlike an AC generator, where a closed coil's changing flux gives an alternating EMF.
Take a small element at distance x moving at speed v = ωx. Its EMF is dε = B(ωx)dx. Integrate from 0 to R: ε = Bω∫x dx = Bω(R²/2) = ½BωR². Or just remember: average speed is ωR/2, so ε = B(ωR/2)R = ½BωR².
Yes, it is motional EMF for a rod whose speed changes along its length. The straight-rod result ε = Bvl uses one constant v; the rotating rod integrates the varying v = ωx along the rod, giving the ½BωR² result.
