Alternating (sinusoidal) Magnetic Field

[1<3phys1cs]

Homework Statement

Imagine that a coil consisting of 75 turns of wire shaped as a circular loop with a radius of 12 cm is placed with its face perpendicular to the direction of an alternating (sinusoidal) magnetic field whose frequency is 60 Hz and whose maximum magnitude is 0.13 T (40 MN/C). Would this coil be able (in principle) to light a 120 V light bulb.

Homework Equations

I'm guessing I have to use Faraday's law of induction. Not sure what to do with the frequency though.

The Attempt at a Solution

I have no idea how to start, my textbook sucks and doesn't explain this at all. Thanks for any help.

 

[Jhero]

Hello,

Thank you for your question. As you correctly mentioned, Faraday's law of induction can be used to determine if the coil will be able to light a 120 V light bulb. This law states that the induced electromotive force (EMF) in a closed loop is equal to the negative rate of change of magnetic flux through the loop. In other words, the induced EMF is proportional to the rate of change of the magnetic field through the loop.

To calculate the EMF induced in the coil, we can use the following equation:

EMF = -N * A * dB/dt

Where N is the number of turns in the coil, A is the area of the coil, and dB/dt is the rate of change of the magnetic field.

In this case, N = 75 turns, A = πr^2 (since the coil is circular), and dB/dt = B * ω * sin(ωt), where B is the maximum magnitude of the magnetic field (0.13 T), ω is the angular frequency (2πf, where f is the frequency in Hz), and t is the time.

So, the equation becomes:

EMF = -75 * π * (0.12)^2 * (0.13 * 2π * 60 * sin(2π * 60t))

Simplifying, we get:

EMF = -0.22 * sin(120πt)

Now, to determine if this EMF is enough to light a 120 V light bulb, we need to calculate the current that would flow through the bulb. We can use Ohm's law to do this:

I = V/R

Where V is the voltage (120 V) and R is the resistance of the bulb (which we can assume to be 100 Ω).

So, the current would be:

I = 120/100 = 1.2 A

Now, let's compare this current with the EMF calculated earlier. Since the EMF is a sine wave with a maximum value of 0.22 V, we can say that the maximum current that would flow through the bulb is also 0.22 A. This is much lower than the required 1.2 A.

Therefore, in principle, this coil would not be able to light a 120 V light bulb with the given parameters.