# mutual inductance

Discussion in 'Homework Help' started by lemon, Feb 13, 2010.

1. ### lemon Thread Starter Member

Jan 28, 2010
125
2
Hi:
a) The flux linking a coil of 2000 turns fall from 0.25Wb to zero in 12.0s. What is the magnitude of the induced emf in the coil?

emf=d(Nδ)/dt=(2000x0.25)/12=41.67v

b) The current in one coil of a pair of coils changes at 0.85A/s. Calculate the induced emf in the other coil if this coil is open circuit and the mutual inductance of the pair is 400mH.

Vs=-M x dip/dt=-400x10^-3 x 0.85 = -0.34V

Would somebody kindly check this please?

2. ### mik3 Senior Member

Feb 4, 2008
4,846
63
You are correct.

3. ### lemon Thread Starter Member

Jan 28, 2010
125
2
yeah!!
I think that is the only time this semester

4. ### lemon Thread Starter Member

Jan 28, 2010
125
2
Thanks for checking mik3
But should the answer to b) -0.34v be a positive number?
Or is it ok that voltage is negative as this is just showing the direction of the induced emf?

5. ### mik3 Senior Member

Feb 4, 2008
4,846
63
It is ok to be negative since it has to cause a current (if possible) opposite to the current creating the magnetic field as to cancel the magnetic field which induced the emf (Lenz's law).