Back emf in a DC Motor plays a role of a regulator,We know the basic equation V=IaRa+Eb,where Ia,Ra are is the current through and resistance of the armature coil,Eb is the back emf and V is the Pd applied.Consider,
Ia=V-Eb/Ra,The value of Ra is really small, just a few ohms and V would generally be around 240-220V(Indian conditions),assuming Eb to be zero(at the time of starting),it shows that Ia comes to be a really large value that the motor coil can barely handle,thats why we use a 3,2 POINT Starter that avoids large starting current,Now in the running condition of the motor Back emf would naturally be produced(say around 200V) this tends to reduce the value of(V-Eb), thus stopping a large current from entering the coil.But as the Load applied to the motor increases , Back emf would naturally come down to allow more current to enter the coil upto its rated value,to deliver the load requirement.thus Eb acts as a regulator.(Eb opposes the applied voltage)

well to be very simple back electro motive force...itself describes wat it means.. d.c motors itself generate an force equivalent to V=I*R+Eb,whre Eb is the BACK E.M.F gereated by armature COIL of motor,actin as an Feedback OPAMP in my ELECTRONICS or as an REGULATOR in electrical terms

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