Physics Asked on May 27, 2021
I understand how the torque depends on the number of turns as
$$ tau = NiAB sin{theta} $$
where $tau$ is the torque, $N$ the number of loops, $i$ the DC intensity, $B$ the magnetic field strength and $theta$ the angle between the coil and the magnetic field.
Therefore, I understand how the torque will increase if we increase the number of loops and keep the rest constant. However, the angular speed $omega$ will also depend on the mass of the coil though the moment of inertia $I$ because
$$tau = NIalpha = NI frac{domega}{dt} $$
where $I$ is the moment of inertia of a single coil. Therefore
$$ frac{domega}{dt} = frac{i A B}{I} sin{theta}$$
So, regardless of the solution to this equation, it should not depend on the number of turns, but here and here it says it depends.
What did I do wrong?
Your mistake is in your conclusion that
$$tau = NIalpha$$
where $I$ is the moment of inertia of a single loop. It's not that simple. The moment of inertia along any particular axis depends on the mass of the body and also on the distribution of that mass relative to that axis.
$$I=int r^2dm$$
where $r$ is the distance of an elemental mass $dm$ from the axis of rotation.
Now consider a current carrying loop in a DC motor. Here the loop rotates around an axis that passes through its plane and is parallel to one of its sides as shown here.
See how the axis of rotation is perpendicular to the screen of the device you are reading this on. Now if we add more loops to this motor, then each loop wont contribute equally to the moment of inertia. The distance of every loop from the axis is not the same.
As $I=int r^2dm$ the contribution of each loop to the moment of inertia dies off as the square of the distance of that loop from the axis. Hence we cannot say that the moment of inertia of the whole body is $N$ times the moment of inertia of one loop.
Answered by Aryamann on May 27, 2021
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