Question

A coil in a magnetic field is removed with:
(i) fast speed,
(ii) slow speed.
In which case, is the induced emf and work done more?

Answer 1

Hint: When we draw the coil fast, the emf produced will be more due to the variation in flux, so it resists the motion. Then we want to give more work to draw. When we pull gradually the flux difference is more minor, the less emf and small work.

Complete step-by-step solution:
The formula for induced emf is given as:
$E = \dfrac{-d \phi}{dt}$
$\phi$ : magnetic flux
t: time
When speed is fast, dt is minimum. And, the dt is inversely proportional to the induced emf. So, the induced emf is more for fast speed.
Work done, $W = P \times t$
The formula for power is given by:
$P = \dfrac{(Blv)^{2}}{r}$
B is the magnetic field.
l is the length.
v is the velocity with which the coil is moving.
$P \propto v^{2}$
When velocity is more, power will be more. With more power, more work will be done.
When we pull the coil fast, there is a change in magnetic flux then the emf induced will be more. So, it opposes the motion. Thus, we need to give more work to pull.
Hence, when a coil in a magnetic field is removed with fast speed, induced emf and work done will be more.

Note:Lenz’s law states that the current generated by the induced emf moves in the direction that resists the change in flux that produced the emf. This is indicated by the negative sign in Faraday’s law. If you take the north end of the magnet towards the coil, the produced emf generates a current whose linked magnetic field has its north end tending towards the magnet and vice versa.