A piece of copper wire is formed into a single circular loop of radius 18 cm. A magnetic field is oriented

3 answers
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

Answers

0.0956431731568 J

Explanation:

N = Number of turns = 1

r = Radius = 18 cm

A = Area = [tex]\pi r^2[/tex]

[tex]\dfrac{dB}{dt}=\dfrac{0.54}{0.57}[/tex]

Induced voltage is given by

[tex]V=NA\dfrac{dB}{dt}\\\Rightarrow V=1\times\pi 0.18^2\times \dfrac{0.54}{0.57}\\\Rightarrow V=0.096430359767\ V[/tex]

Resistance is

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

Power is given by

[tex]P=\dfrac{V^2}{R}\\\Rightarrow P=\dfrac{0.096430359767^2}{0.0554176944093}\\\Rightarrow P=0.167795040626\ W[/tex]

Energy

[tex]E=Pt\\\Rightarrow E=0.167795040626\times 0.57\\\Rightarrow E=0.0956431731568\ J[/tex]

The average electrical energy dissipated is 0.0956431731568 J

Answer: fn = wcos50 = 38.89(9.81)(cos50) = 245.23 n

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