Recent content by in10s3gamer

Okay, 100% clear now, ready to be locked

Something was wrong with my calculator, and now I get the right answer. However, why is there only one factor of 0.016 in the calculation for flux? Isn't flux = BA, and A is equal to 0.016^2.

If I take the derivative of the flux with respect to t i get your equation for flux, but instead of (C+100t) it is just (100). After calculating the derivative of the flux with respect to t, this equals the EMF. I then set EMF = IR, and so the current = EMF/R. Since R = 1.1*10^-2, I get...

Where does the 0.016 come from? edit: i think 0.016 should be 0.016^2

Thank you, it is very clear now. edit: r is in the denominator, so do i end up with [100mu/(2pi)]ln(r), with r evaluated from 0.01 - 0.026?

That's where I was stuck earlier. I'm not exactly sure what I is equal to. I am thinking it is a function of time. Is it 100t, where t equals time in seconds? Or do I have to include the current inside of the square loop as well?

The flux's sign is arbitrary, but it's equal to BA, which is B(0.016)^2 in this case. And B can be determined by Biot-Savart's law? Integrating from 0 to 0.016: \frac{\mu_0}{4\pi}\times \frac{I}{r^3}\int dl Is that correct? From that I would get B, then I could plug in the function for the...

Homework Statement A 1.60 cm x 1.60 cm square loop of wire with resistance 1.10 \times 10^{-2} \Omega is parallel to a long straight wire. The near edge of the loop is 1.10 cm from the wire. The current in the wire is increasing at the rate of 100 A/s Homework Equations I think I have...