Emf max for a guitar string.

[Ithryndil]

Homework Statement

Review problem. A flexible metallic wire with linear density 3.00 multiplied by 10-3 kg/m is stretched between two fixed clamps 64.0 cm apart and held under tension 301 N. A magnet is placed near the wire as shown in Figure P31.25. Assume that the magnet produces a uniform field of 5.00 mT over a 2.00 cm length at the center of the wire and a negligible field elsewhere. The wire is set vibrating at its fundamental (lowest) frequency. The section of the wire in the magnetic field moves with a uniform amplitude of 1.50 cm.

http://www.webassign.net/serpop/p23-18.gif

(b) Find the amplitude of the electromotive force (emfmax) induced between the ends of the wire.

Homework Equations

Emf = BA

The Attempt at a Solution

I am not quite sure I am doing the problem correctly. I had the area being 3.00cm X 2.00cm but that's not correct nor do I actually believe the above equation is the correct equation. I assume the equation needs to have a sine or a cosine in it, but we are needing to find Emfmax so that's not too important right? Thanks for the help.

 

[Redbelly98]

Correction to your equation:

$$Emf = \frac{d}{dt} (BA)$$

Since B is constant, the problem comes down to finding dA/dt.

 

[thomate1]

I would approach this problem by first understanding the physical principles involved. In this case, we have a wire moving in a magnetic field, which will induce an emf according to Faraday's law of induction. The equation you stated, Emf = BA, is actually the correct equation, where B is the magnetic field and A is the area of the wire perpendicular to the field.

To find the maximum emf, we need to consider the maximum area of the wire that is moving in the magnetic field. In this case, it is the section of the wire that is 2.00 cm long and has an amplitude of 1.50 cm. This gives us an area of 2.00 cm x 1.50 cm = 3.00 cm^2.

Now, we need to convert this to meters since the linear density and tension were given in SI units. 3.00 cm^2 is equivalent to 3.00 x 10^-4 m^2.

Finally, we can plug this into the equation Emf = BA, along with the given magnetic field of 5.00 mT, to get:

Emfmax = (5.00 x 10^-3 T)(3.00 x 10^-4 m^2) = 1.50 x 10^-6 V

Therefore, the maximum emf induced between the ends of the wire is 1.50 microvolts. I would also recommend double checking the units to make sure they are consistent throughout the calculation.