Question

If 55.0 cm of copper wire (diameter = 1.00 mm) is formed into a circular loop and placed perpendicular to a uniform magnetic field that is increasing at a constant rate of 11.5 mT/s, at what rate is thermal energy generated in the loop?

Answer 1

We find the area of the wire loop first.

$$0.55m = 2\pi r$$ $$r = \frac{0.55m}{2\pi} = 8.75 \cdot 10^{-2}m$$ $$A_{loop} = \pi r^2 = 2.4 \cdot 10^{-2}m^2$$

Now that we know the area, we can compute the change in flux and induced emf.

$$|\varepsilon| = \left| \frac{d\varphi_B}{dt} \right| = \left| \frac{d}{dt} (BA_{loop}) \right| = A_{loop} \frac{dB}{dt} = 2.8 \cdot 10^{-4}Tm^2$$

The power will depend on the resistance of the wire.

$$R = \rho_{Cu} \frac{L}{A_{wire}} = 1.2 \cdot 10^{-2}\Omega$$ $$P = \frac{\varepsilon^2}{R} = 6.5 \cdot 10^{-6}W$$