Question

A square conducting loop with sides of length \(L\) is rotating at a constant angular speed, \(\omega\), in a uniform magnetic field of magnitude \(B\). At time \(t=0,\) the loop is oriented so that the direction normal to the loop is aligned with the magnetic field. Find an expression for the potential difference induced in the loop as a function of time.

Short answer

Answer: The potential difference induced in the square conducting loop as a function of time is given by the expression: $$ V(t) = - B \cdot L^2 \cdot \omega \cdot \sin(\omega t). $$

Step-by-step solution

Identify the magnetic flux through the loop

The magnetic flux through the loop, \(\Phi\), is given by the product of the magnetic field, \(B\), the area of the loop, \(A\), and the cosine of the angle between the magnetic field and the normal to the plane of the loop, \(\theta\). The area of the square loop is \(A=L^2\), where \(L\) is the side length, and the angle \(\theta\) between the magnetic field and the normal to the plane of the loop changes with time as the loop rotates at a constant angular speed \(\omega\). So, the magnetic flux through the loop is: $$ \Phi(t) = B \cdot A \cdot \cos(\theta) = B \cdot L^2 \cdot \cos(\theta(t)). $$

Find \(\theta(t)\) in terms of \(\omega\) and \(t\)

The loop is rotating at a constant angular speed \(\omega\). So, the angle between the magnetic field and the normal to the plane of the loop can be expressed as a function of time as: $$ \theta(t) = \omega t. $$ Now, we can rewrite the magnetic flux through the loop as a function of time: $$ \Phi(t) = B \cdot L^2 \cdot \cos(\omega t). $$

Apply Faraday's law of electromagnetic induction

Faraday's law states that the induced electromotive force (EMF) in a closed loop is equal to the rate of change of the magnetic flux through the loop: $$ \varepsilon(t) = -\frac{d\Phi(t)}{dt}. $$ Let's differentiate the magnetic flux with respect to time \(t\): $$ \frac{d\Phi(t)}{dt} = - B \cdot L^2 \cdot \sin(\omega t) \cdot \omega. $$

Find the potential difference induced in the loop

Now, we can write the expression for the induced potential difference, which is equal to the induced EMF, as a function of time: $$ V(t) = \varepsilon(t) = - B \cdot L^2 \cdot \omega \cdot \sin(\omega t). $$

Final answer

The potential difference induced in the square conducting loop as a function of time is: $$ V(t) = - B \cdot L^2 \cdot \omega \cdot \sin(\omega t). $$