Using Ohm's Law to Confirm the Relationship Between Loop Radius and Peak Voltage

[lampshader]

Homework Statement

Using Ohm’s Law as a check: V = IR, where V = emf in loop, what is the ratio of the current in the loop of radius 20 cm compared to the loop of radius 40 cm?

(It should be the same as the ratio of the emfs!) Treat the loop as a solid circle and remember the formula for the area of a circle is A = π r2. So if the radius is halved, what does this do to the area? Does this agree with the ratio of the two peak voltages that you calculated? Record your observations and explain.

Homework Equations

V = IR,

m = (y2-y1) / (x2-x1)

The Attempt at a Solution

I gathered all my data (which is the first part of the question):

Table 1:
Loop Radius (cm) Peak Voltage (V)
10 0.138
20 0.553
30 1.244
40 2.211
50 3.454

The data in table 1 is suggesting that as the radius increases, so will the peak voltage.

So my instructor wants me to check it using Ohm's Law, How exactly would I do that? Here is the question again:

Using Ohm’s Law as a check: V = IR, where V = emf in loop, what is the ratio of the current in the loop of radius 20 cm compared to the loop of radius 40 cm?

We have:
I = V / R = 0.553 / 20 = 0.27650;
And:
I = V / R = 2.211 / 40 = 0.55275;

My instructor is saying that it should be the same as the ratio of the emf's

 

[jbsteele]

. So my ratio would be: Ratio = 0.27650 / 0.55275 = 0.5. It does agree with the ratio of the two peak voltages that I calculated. The ratio of the peak voltages is 0.5, which is exactly the same as the ratio of the currents. This means that Ohm's Law is valid in this case, as it confirms that increasing the radius of the loop increases the peak voltage, which is exactly what I observed from the data in Table 1.