Question

If 62.9 cm of copper wire (diameter = 1.15 mm, resistivity = 1.69 × 10-8Ω·m) is formed into a circular loop and placed perpendicular to a uniform magnetic field that is increasing at the constant rate of 8.43 mT/s, at what rate is thermal energy generated in the loop?

Answer 1

Answer:

The answer is "$\bold{7.30 \times 10^{-6}}$"

Explanation:

length of the copper wire:

L= 62.9 cm

r is the radius of the loop then:

$r=\frac{L}{2 \pi}$

  $=\frac{62.9}{2\times 3.14}\\\\=\frac{62.9}{6.28}\\\\=10.01$

area of the loop Is:

$A_L= \pi r^2$

     $=100.2001\times 3.14\\\\=314.628$

change in magnetic field is:

$=\frac{dB}{dt} \\\\ = 0.01\ \frac{T}{s}$

then the induced emf is:  $e = A_L \times \frac{dB}{dt}$

                                              $=314.628 \times 0.01\\\\=3.14\times 10^{-5}V$

resistivity of the copper wire is: $\rho =$  1.69 × 10-8Ω·m

diameter d = 1.15mm

radius (r) = 0.5mm

               $= 0.5 \times 10^{-3} \ m$

hence the resistance of the wire is:

$R=\frac{\rho L}{\pi r^2}$

   $=\frac{1.69 \times 10^{-8}(62.9)}{3.14 \times (0.5 \times 10^{-3})^2}\\\\=\frac{1.69 \times 10^{-8}(62.9)}{3.14 \times 0.5 \times 0.5 \times 10^{-6}}\\\\=\frac{1.69 \times 10^{-8}(62.9)}{3.14 \times 0.25 \times 10^{-6}}\\\\=135.41 \times 10^{-2}\\=1.35\times 10^{-4}$

Power:

$P=\frac{e^2}{R}$

$=\frac{3.14\times 10^{-5}\times 3.14\times 10^{-5}}{1.35 \times 10^{-4}}\\\\=7.30 \times 10^{-6}$

The final answer is: $\boxed{7.30 \times 10^{-6} \ W}$