Question

A piece of copper wire is formed into a single circular loop of radius 18 cm. A magnetic field is oriented parallel to the normal to the loop, and it increases from 0 to 0.54 T in a time of 0.57 s. The wire has a resistance per unit length of 4.9 x 10-2 /m. What is the average electrical energy dissipated in the resistance of the wire

Answer 1

Answer:

0.0956431731568 J

Explanation:

N = Number of turns = 1

r = Radius = 18 cm

A = Area = $\pi r^2$

$\dfrac{dB}{dt}=\dfrac{0.54}{0.57}$

Induced voltage is given by

$V=NA\dfrac{dB}{dt}\\\Rightarrow V=1\times\pi 0.18^2\times \dfrac{0.54}{0.57}\\\Rightarrow V=0.096430359767\ V$

Resistance is

$R=2\pi rR/m\\\Rightarrow R=2\pi 0.18\times 4.9\times 10^{-2}\\\Rightarrow R=0.0554176944093\ \Omega$

Power is given by

$P=\dfrac{V^2}{R}\\\Rightarrow P=\dfrac{0.096430359767^2}{0.0554176944093}\\\Rightarrow P=0.167795040626\ W$

Energy

$E=Pt\\\Rightarrow E=0.167795040626\times 0.57\\\Rightarrow E=0.0956431731568\ J$

The average electrical energy dissipated is 0.0956431731568 J