Question

A square loop (side a) is mounted on a vertical shaft and rotated at angular velocity $\omega$ (Fig. 7.19). A uniform magnetic field B points to the right. Find the$\epsilon \left(t\right)$for this alternating current generator.

Short answer

The induced emf in the square loop is .$B{a}^2\omega \sin \omega t$

Step-by-step solution

Write the given data from the question.

The uniform magnetic field is B .

The side of the square loop is a .

The angular velocity is $\omega$.

Calculate the generated emf in the square loop.

$\epsilon \left(t\right)=-\frac{d}{dt}\left(BA\cos \omega t\right)$The area of square loop,$A=a^2$ .

The square loop is moving at angle with angular velocity in time

$t\theta \omega t$.

The magnetic flus through the loop is given by,

role="math" localid="1657618600560" $\varphi =BA\cos \theta$

Substitute $\omega t$for $\theta$into above equation.

$\varphi =BA\cos \theta$

According to the Faraday’s law, the induced emf in any closed loop is equal to the negative of the rate of change of flux in the circuit.

$\mathit{\epsilon }\left(t\right)\mathbf{=}\mathbf{-}\frac{\mathbf{d}\mathbf{\varphi }}{\mathbf{d}\mathbf{t}}$

Substitute $\varphi =BA\cos \omega t$for $\varphi$into above equation.

$\epsilon \left(t\right)=-\frac{d}{dt}\left(BA\cos \omega t\right)$

Substitute for into above equation.

role="math" localid="1657619049126" $$\begin{gathered} \epsilon \left(t\right)=-\frac{d}{dt}\left(BA\cos \omega t\right) \\ \epsilon \left(t\right)=-B{a}^2\left(-\sin \omega t\right).\omega \\ \epsilon \left(t\right)=-B{a}^2\sin \omega t \end{gathered}$$

Hence the induced emf in the square loop is $B{a}^2\omega \sin \omega t$.