Consider a cylindrical magnet and a solenoid (made of copper wire) in the same horizontal line. The magnet is to the left of the solenoid at some distance ($S$). The magnet is moving from left to right with a constant velocity and after some time $(t)$ it will enter the solenoid.
Let the parameters of the setup be:
$m_0,v_0,d_0,l_0$=mass,velocity,diameter and length of the magnet
$m_1,d_1,l_1$=mass,inner diameter (hollow) and length of the solenoid
$d_0<d_1$ such that magnet passes though solenoid without any physical contact
As the magnet passes through the solenoid an EMF would be generated that would oppose the motion of magnet. That emf would decrease the velocity of the magnet. this decrease of velocity would in turn decrease the emf. And this feedback loop will continue till the magnet comes to a halt.
The equation of the EMF in given by: $∈=-(d(∅_BNπr^2 cos(θ))/dt$
Since the induced EMF is proportional to the rate of change of magnetic flux. And (if I am correct) the rate of change of flux is proportional to the velocity of the magnet through the solenoid. The EMF opposes the motion of the magnet through the solenoid thus the magnet’s velocity through the coil decreases with time. This in turn decreases the EMF with time.
Case 1: The solenoid is attached to ground thus it doesn't move.
Case 2: The solenoid is initially at rest but is allowed to move from left to right.
Query: How would the plot of decrease in velocity and decrease in emf look like with time for any velocity $v_0$ (we are assuming a long enough solenoid and despite the solenoid being attached to ground no external forces like gravity etc) in these cases?
Can anyone please help with the plots (velocity vs time and EMF vs time)?
Best Answer
For the purpose of force analysis, the magnet can be treated as another solenoid with a fixed current flowing around the outer surface (giving an equivalent dipole moment). Forces occur near the ends of the outer solenoid where it's induced field has radial components.As the leading end of the magnet approaches and enters the solenoid, the increasing flux produces an emf in nearby turns of the solenoid. This can cause a current in the solenoid circuit, and the spreading field at the end of the solenoid will cause a force which slows the magnet. As the trailing end of the magnet approaches and enters the solenoid, it will cause an emf which opposes that produced by the leading end. The current and force disappear. As the leading end leaves the far end of the the solenoid, its effect fades. The current in, and field from, the solenoid is reversed. But the spread of the field at the far end is also reversed. Again the force on the magnet causes it to slow. In both cases K.E. from the moving magnet is dissipated as heat in the circuit of the solenoid. As the trailing end of the magnet moves away from the solenoid, all such effects disappear.