# mutual inductance

#### lemon

Joined Jan 28, 2010
125
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? #### mik3

Joined Feb 4, 2008
4,843
You are correct. #### lemon

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

#### lemon

Joined Jan 28, 2010
125
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?

#### mik3

Joined Feb 4, 2008
4,843
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?
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).