A motor has a single loop inside a magnetic field of magnitude \(0.87 \mathrm{~T}\). If the area of the loop is \(300 \mathrm{~cm}^{2}\), find the maximum angular speed possible for this motor when connected to a source of emf providing \(170 \mathrm{~V}\).

To start, we must determine the magnetic flux \(\phi\) in the loop. The magnetic flux is given by the product of the magnetic field \(B\), the loop area \(A\), and the cosine of the angle between the magnetic field and the normal to the loop surface, which can be represented as: \(\phi = B \cdot A \cdot \cos{\theta}\) Since the motor has a single loop, we know that at its maximum angular speed, the magnetic field and area vector will be perpendicular to each other. Therefore, the angle between them will be \(90^\circ\), and the cosine of \(90^\circ\) is 0. So, the magnetic flux is given by: \(\phi = B \cdot A \cdot \cos{90^\circ} = 0.87\ (\mathrm{T}) \cdot 300\ (\mathrm{cm}^{2}) \cdot 0 = 0\ \mathrm{Wb}\)

Now, we apply Faraday's Law, which states that the induced emf in the loop is equal to the rate at which the magnetic flux changes. Faraday's Law is given by: \(|ε| = N\ |\frac{d\phi}{dt}|\) Since the motor has only one loop, \(N = 1\). Also, we are given the source emf as \(170\ \mathrm{V}\) (it is assumed that this is equal to the induced emf). Therefore, we can rewrite Faraday's Law for our problem as: \(|ε| = |\frac{d\phi}{dt}|\)

We are now looking for the maximum angular speed ω of the motor. Recall the magnetic flux expression: \(\phi = B \cdot A \cdot \cos{\theta}\) We know that \(\theta = \omega t\). So: \(\phi = B \cdot A \cdot \cos{\omega t}\) We then find the derivative of \(\phi\) with respect to time \(t\): \(\frac{d\phi}{dt}=-\omega B A \sin{\omega t}\) Substitute the derivative back into Faraday's Law and solve for the angular velocity ω: \(|ε| = |-ω B A \sin{\omega t}|\) \(\frac{170 \mathrm{\ V}}{0.87\ (\mathrm{T})\cdot 300\ (\mathrm{cm}^2)} = \sin{\omega t}\) Evaluate the sine function to find the maximum angular speed: \(\omega = \frac{1}{t} \arcsin{\left(\frac{170\ \mathrm{V}}{0.87\ (\mathrm{T})\cdot 300\ (\mathrm{cm}^2)}\right)}\) Notice that to obtain the maximum value of \(\omega\), the sine function must be at its maximum value, which is 1. This occurs when the argument inside the arcsin function is 1, so: \(\frac{170\ \mathrm{V}}{0.87\ (\mathrm{T})\cdot 300\ (\mathrm{cm}^2)} = 1\) Finally, solve for \(\omega\): \(\omega = \frac{170\ \mathrm{V}}{0.87\ (\mathrm{T})\cdot 300\ (\mathrm{cm}^2)}\) \(\omega \approx 0.65\ \mathrm{rad/s}\)

The maximum angular speed possible for this motor when connected to a source of emf providing \(170 \mathrm{\ V}\) is approximately \(0.65\ \mathrm{rad/s}\).