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, we can use Faraday's Law of electromagnetic induction. According to Faraday's Law, the magnitude of the induced electromotive force (emf) is equal to the rate of change of the magnetic flux through the loop.
The magnetic flux through the loop is given by the product of the magnetic field strength (B), the area of the loop (A), and the cosine of the angle between the normal to the loop and the magnetic field. In this case, the angle between the normal to the loop and the magnetic field is changing with time due to the rotation of the loop. Let's call this angle 'theta'.
First, let's find the expression for the area of the loop. Since it is a square loop, each side has a length L, so the area (A) of the loop is given by A = L^2.
Now, the flux through the loop can be expressed as Phi = B * A * cos(theta), where B is the magnetic field strength and A is the area of the loop.
As the loop is rotating at a constant angular speed, the angle theta changes with time, and it can be expressed as theta = wt, where w is the angular speed and t is time.
Now, let's calculate the rate of change of flux with respect to time, d(Phi)/dt:
d(Phi)/dt = d/dt (B * A * cos(theta))
= B * A * d(cos(theta))/dt
= -B * A * sin(theta) * d(theta)/dt
Since d(theta)/dt = w (since angular speed is constant), we can simplify the expression for d(Phi)/dt as:
d(Phi)/dt = -B * A * w * sin(theta)
Now, according to Faraday's Law, the magnitude of the induced emf is equal to the absolute value of the rate of change of the magnetic flux through the loop. So, the potential difference induced in the loop, V_ind, can be expressed as:
V_ind = |d(Phi)/dt|
= B * A * w * |sin(theta)|
Since sin(theta) is always positive, we can rewrite the expression as:
V_ind = B * A * w * sin(theta)
Plugging in the expression for the area (A = L^2), the final expression for the potential difference induced in the loop as a function of time is:
V_ind = B * L^2 * w * sin(theta)
So, you were correct! The expression for the potential difference induced in the loop is indeed V_ind = B * L^2 * w * sin(theta).