Say we have a motor coil like this: enter image description here

We hang a mass (red ball) on the motor to prevent it's rotation. We make the mass heavy enough such that it's Weight Force directly opposes the motor force produced by that wire.

$$mg = BIL$$

Does this motor turn? I feel like the answer is no, because that wire has no net force acting on it (forces cancel out). However, there still is a force being produced by the right hand wire (next to the N pole). It feels like this force should still be able to make the coil turn.


Answer: Yes, motor's coil will turn.

Notice, the magnetic field $B$ exerts a force $=BIL$ to the right hand wire in vertically downward direction (given by Fleming left hand rule). Similarly, it exerts an equal force $=BIL$ on left hand wire in vertically upward direction. These two equal and opposite forces for a couple which have tendency to turn the coil of motor depending on the magnitude of net turning moment. $$\text{Turning moment acting on the coil}=BIL\times d$$ $$\text{Opposing moment prouced by weight}, (mg=BIL)=mg\times \frac{d}{2}=BIL\times \frac d2$$ $$\implies BIL\times d>mg\times \frac{d}{2}$$ Since, turning moment is greater than opposing moment by weight $mg$ Hence the coil will certainly turn.


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