A 2.40 m length of wire is held in an east-west direction and moves horizontally to the north with a speed of 15.1 m/s. The vertical component of Earth's magnetic field in this region is 40.0 µT directed downward. Calculate the induced emf between the ends of the wire and determine which end is positive.
*I tried using the formula E=Blv where B is the magnetic field (4.0e-5 T), but the program says I'm off by a multiple of ten. I'm assuming I'm making a stupid mistake, but I can't figure out what it is!
11 years ago ? Interesting
BLv is correct. 40E-6*2.40*15.1=1.45mVolts I get.
To calculate the induced electromotive force (emf) in this situation, you correctly started with the formula E = Blv, where E is the induced emf, B is the magnetic field strength, l is the length of the wire perpendicular to the magnetic field, and v is the velocity of the wire.
Here are the steps to solve this problem:
1. First, convert the given magnetic field strength from microtesla (µT) to tesla (T). Since 1 T = 10^6 µT, the magnetic field strength is 40.0 µT = 40.0 × 10^(-6) T = 4.0 × 10^(-5) T. So, your value for B is correct.
2. Next, consider the length of the wire (l) in the formula. In this case, the wire is moving horizontally to the north, so only the vertical component of the length is relevant. Given that the wire is 2.40 m long and held in an east-west direction, you need to consider the vertical component of the wire's length. Using trigonometry, you can find that the vertical component of the length is l = 2.40 m * sin(90°) = 2.40 m.
3. Now, you have the magnetic field strength (B = 4.0 × 10^(-5) T), the length of the wire perpendicular to the magnetic field (l = 2.40 m), and the velocity of the wire (v = 15.1 m/s). Plug these values into the formula E = Blv:
E = (4.0 × 10^(-5) T) * (2.40 m) * (15.1 m/s)
E = 0.00288 V
Therefore, the induced emf between the ends of the wire is 0.00288 volts.
4. To determine which end of the wire is positive, consider the direction of the induced current. By convention, the positive end of the wire is the one from which current will flow when it is induced. In this case, since the wire is moving horizontally to the north and the magnetic field is directed downward, the induced current will flow counterclockwise. Therefore, the end of the wire moving to the north is positive and the end moving to the south is negative.
So, the positive end of the wire is the one moving horizontally to the north.