Magnetic Flux Through a Wire Coil

[cepheid]

Hi,

This problem confused me the first time around, but I think I've got it. I just want a confirmation that my reasoning is correct. I know I have the correct answer, but what is important is the method of arriving at it.

Homework Statement

A wire is being wound around a rotating wooden cylinder of radius R. One end of the wire is connected to the axis of the cylinder. The cylinder is placed in a uniform magnetic field of magnitude B parallel to its axis and rotates at N revolutions per second. What is the potential difference between the two open ends of the wire?

Homework Equations

$$\oint \mathbf{E} \cdot d\mathbf{l} = - \frac{d}{dt}\int_A \mathbf{B} \cdot \mathbf{\hat{n}} \, dA$$
​

which can be written as

$$\textrm{emf} = -\frac{d\Phi_B}{dt}$$
​

The Attempt at a Solution

Here is my reasoning, which I want checked:

The potential difference is determined by the rate of change of magnetic flux through the coil in between the two ends of the wire. Since this coil keeps growing, the flux is indeed changing. Each turn of the coil has magnetic flux $B \pi R^2$ through it, and each revolution of the cylinder adds a turn. There are N revolutions per second, which means that the coil increases by N turns per second. The flux therefore increases by $NB \pi R^2$ every second; this is its rate of change and is therefore equal (at least in magnitude) to the potential difference between the two ends of the wire.

 

[marcusl]

This looks correct to me.

 

[m2003]

Your reasoning is correct. The potential difference between the two ends of the wire is equal to the rate of change of magnetic flux through the coil. As the coil rotates and grows, the magnetic flux through it also changes, resulting in a potential difference between the two ends of the wire. Your equation, emf = -dΦB/dt, accurately represents this relationship. Well done!