1. Inductance

2. Mutual Inductance First we discuss mutual inductance
Suppose we have 2 coils. Here is 1 coil and a 2nd one is near by.
Now suppose we pass a current I(t) that is time dependant through the 1st coil. As I is time dependant, it will produce a magnetic field B that is also time dependant.
This B field will produce a flux in the 2nd coil which is also time dependant.
Now by Faraday’s law, if we have a time dependant flux, we have an induced EMF in the 2nd coil.
This induced EMF is –dΦ/dt
The flux Φ, depends on B. If B doubles, so does Φ, and B is proportional to I.
So we find that Φ in the 2nd coil is proportional to I in the 1st coil.
Φ2 = MI1 where M is the mutual inductance.
E = -MdI1/dt
Now notice that M depends only on the geometry, like what type of coil, the size, the number of turns etc.
Suppose I2 is in the 2nd coil. What is the fluz through the first coil?
Φ1 = MI2 it’s the same M
Units = Henry = T.m2/A = H

3. Example Let’s suppose I have 2 coils and the M= 0.001H = 1mH
And suppose the current in the 1st coil I1 = 10 sin(120πt) A, where t is in seconds
First we’d like to know, what is Φ in coil 2 at t=0
Induced EMF in coil 2 at t=0
To answer this we know that Φ2 = MI1 =0.01sin(120πt) = 0
One may think that since the flux is 0, then the EMF is zero.
This is not true.
A function can be zero at a point, but it’s derivative won’t be.
E = -dΦ/dt = -120π(0.01)cos (120πt) V
So at t=0, E = -1.2 pi V
The minus sign indicates the direction, so the induced emf is about 4 volts.

4. Example Let’s say we have a solenoid with n turns/unit length
And suppose I place a ring inside the solenoid such that the plane of the ring is perpendicular to the axis of the solenoid.
Now we pass a current through the solenoid and we ask for the flux through the small ring.
Let’s suppose the radius of the solenoid is R and the small ring is a.
So we want to find the flux through the small ring and the mutual indicatance.
Now with the current through the solenoid there is a constant magnetic field that’s uniform, assuning the solenoid is very long compared to the radius.
B inside solenoid = μ0 nI, where I is current
By the right hand rule it gives us the B field to the left.
Φ through the ring = BA (Area of the ring) = Bπa2 = μ0 nIπa2
Φ = MI => M = μ0 nπa2
Notice that this system M depends on the number of turns and radius (i.e. the geometry)

5. Self Inductance
Now suppose I only have 1 coil and I pass a current through it. This current is time dependant. Again as it’s time dependant a B field that is time dependant also.
Flux Φ through the coil. (time dependant also)
Induced EMF in the coil
Φ depends on the current. Φ proportional to B and I
Φ=LI L= self Inductance again Henrys
E=-LdI/dt we have a pd equal to this
Sometimes this is called the back emf and we see this in circuits all the time. It prevents the current from changing too quickly.
If it changes too quickly. If the current changes sharply, we get a large back emf.

6. Homework Please complete the handout on Inductance
See homework sheet number 19