A square conducting loop with sides of length L is rotation at a constant angular speed of, w (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 the expression for the potential difference induced in the loop as a function of time.

I think it's Vind= BL^2*w*sin(theta)

Is this right? If it's not, can someone please explain how it should be? Thanks.

To find the expression for the potential difference induced in the loop as a function of time, we can use Faraday's Law of Electromagnetic Induction, which 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.

The magnetic flux (Φ) through a loop of area A in a uniform magnetic field B is given by the equation Φ = B*A*cos(θ), where θ is the angle between the direction normal to the loop and the magnetic field. In this case, since the loop is initially aligned with the magnetic field, θ remains constant at 0 degrees.

Since the loop is rotating at a constant angular speed ω, the angle θ between the normal to the loop and the magnetic field changes with time. At any given time t, we can express the angle θ as ω*t.

Using the equation for magnetic flux and knowing that A = L^2 (since the loop is square with sides of length L), we can express the magnetic flux through the loop as Φ = B*L^2*cos(ω*t).

Now, applying Faraday's Law, the induced emf (ε) in the loop is the negative rate of change of the magnetic flux with respect to time. Therefore, we have ε = -dΦ/dt.

Differentiating Φ with respect to time gives us dΦ/dt = -B*L^2*sin(ω*t)*ω.

Finally, the induced potential difference (Vind) across the loop is given by Vind = ε = -dΦ/dt = B*L^2*sin(ω*t)*ω.

So, the correct expression for the potential difference induced in the loop as a function of time is Vind = B*L^2*ω*sin(ω*t).