# Faraday's Law of Induction - Experiment Giving Weird Results?

1. Sep 14, 2016

### Peter2

• Member advised to use the homework template for posts in the homework sections of PF.
1. The problem statement, all variables and given/known data

Hi everyone, some school mates sugested id try this forum for help with this work i have :)
- will try to keep it short

We were trying to induce current in a coil and confirm the experimental results with math, but the math is way off and nobody is sure why...

Experiment:
• Cooper Coil (3Ω) (measured with multimeter)
• Rectangle (6.5 cm x 3,5 cm) → Area ≈ 0.002275 m2
• Number of Turns 6
• Magnet Rectangle (4cm x 10cm x 5cm) (3050 Gauss) → 0.305 Tesla
• Δt of passing the magnet was around 1sec... but we had a hard time keeping this scientific and precise
Results:
• Digital Multimeter set to 2mA - Average of the readings turned out 0.0048
• 0.0048mA → 48x10-7 A
• I = V/A → V = 0.000014 Volts

2. Relevant equations

Now for the Theoric Rusults:

• Φ = B ⋅ A ⋅ cosθ → Φ= 0.305 ⋅ 0.002275 ⋅ cos0 = 0.000694
• ξ= - N ⋅ (dΦ/dt) → ξ = - 6 ⋅ 0.000694/1 = 0.004163 Volts

3. The attempt at a solution

Why there is such a big difference in our math vs our experience?
Are we doing something wrong?

Thanks for any help ;) Cheers!

Last edited: Sep 14, 2016
2. Sep 14, 2016

The equations are simple enough and you are only doing estimates, but if you got reasonably good measurements of the resistance and current, I think you should get results that are reasonably consistent with the theoretical calculations. One thing that inexpensive multimeters often have is an internal resistance that affects the measurements. I do think the resistance of your copper wire is likely to be considerably less than 3 ohms, and if you got a good reading of the current in the circuit without the meter's resistance supplying some impedance to the current flow, I do think the current may have been higher than what you measured. I would recommend using an op-amp based current amplifier (current to voltage amplifier) circuit and measure the output of that circuit with an oscilloscope. Alternatively, you could measure the EMF produced by connecting the coils to the high impedance (1 Mohm) input of the oscilloscope and simply measure the voltage generated from the process of moving the magnet through the coils.

3. Sep 15, 2016

### Peter2

Many thanks for the reply Charles.

Cool, i was worried something was off with the equations, ill add an op - amp and retry the experiment, and maybe even try a new coil with more (30-40) loops to try and compensate for the multimeters internal resistance

4. Sep 15, 2016

Copper wire I believe is quite low resistance. Much of your resistance could occur at the contact point, etc. Perhaps your best EMF measurement would be done directly by reading the voltage from the coils on an oscilloscope. Extra loops to give you a higher voltage would probably improve the accuracy.

5. Sep 15, 2016

### CWatters

I agree with Charles. 3 Ohms sounds high for just 6 turns of wire.

6. Sep 15, 2016

### Staff: Mentor

It might be interesting to investigate what gauge of wire would be required to produce the reported coil resistance of 3 Ω.

The perimeter of the 6.5 cm x 3.5 cm rectangular coil is 20 cm. With 6 turns that yields a total wire length (excluding any connection leads) of 1.2 m. So the wire resistance per meter would be 2.5 Ω/m (or 0.762 Ω/ft for fans of feet and inches). That's close to a 39 gauge copper wire, with a diameter a bit less than a hair's width (less than 0.1 mm). That's a pretty fine wire.

Perhaps the OP can share what he observed regarding the thickness of the coil wire? Did it appear to be as thin as a hair?

An alternative is that the wire is not copper but some other material with more ohms-per-meter. For example, a 20 gauge nichrome wire might fit the bill.

7. Sep 16, 2016

### rude man

You milliammeter is probably much too slow to record the current accurately. The current will come & go in under 1 sec. Most digital m/a meters need several seconds to stabilize the reading.

Use a current-sampling resistor (say 1K ohms in series with the coil & look at the voltage across it with a 'scope. Voltage should peak around 4 mV and current around 4mV/1K = 4 μA.

Last edited by a moderator: Sep 16, 2016
8. Sep 18, 2016

### rude man

Also, you may be generating a counter B field by the loop current sufficient to reduce the vector sum of your applied magnet B field
and the counter-B field to reduce the current appreciably below computed numbers. emf = d/dt A(external B - induced B). The induced effective B as a function of current is very difficult to compute for a thin loop. But using a 1K resistor as I proposed reduces the induced B field to practically zero.