Question

The armature of an ac generator is a rectangular coil \(2.0 \mathrm{cm}\) by \(6.0 \mathrm{cm}\) with 80 turns. It is immersed in a uniform magnetic field of magnitude 0.45 T. If the amplitude of the emf in the coil is $17.0 \mathrm{V},$ at what angular speed is the armature rotating?

Short answer

Answer: The angular speed of the rectangular coil is approximately \(393.52\ \mathrm{rad/s}\).

Step-by-step solution

Write down the formula

We use the formula for the amplitude of the emf in a rotating coil: \(emf = NBA\omega\sin{\omega t}\). Since we are given the amplitude of the emf, we can set \(emf = NBA\omega\).

Substitute the given values

We are given the following values: \(emf = 17.0\ \mathrm{V}\) \(N = 80\) turns \(B = 0.45\ \mathrm{T}\) \(A = 2.0\,\mathrm{cm} \times 6.0\,\mathrm{cm} = 12.0\,\mathrm{cm}^2 = 0.0012\,\mathrm{m}^2\) Now we can plug these values into the equation: \(17.0\ \mathrm{V} = (80)(0.45 \mathrm{T})(0.0012\ \mathrm{m}^2)\omega\)

Solve for the angular speed \(\omega\)

Now we solve for \(\omega\): $$ \omega = \frac{17.0\ \mathrm{V}}{(80)(0.45 \mathrm{T})(0.0012\ \mathrm{m}^2)} = \frac{17.0}{0.0432} \mathrm{rad/s} $$ $$ \omega \approx 393.52\ \mathrm{rad/s} $$

Write the final answer

The angular speed at which the armature rotating is approximately \(393.52\ \mathrm{rad/s}\).