A copper wire of diameter 0.04 in. and length 50 cm is bent in the form of a circular loop. The plane of the loop is normal to a uniform magnetic field which is increasing with time at a constant rate of 100 G `s^(-1)`. What is the rate of joule heating in the loop? [Resistivity of copper `=1.7xx10^(-8)Omega.m, 1" in."=2.54cm`]

Explanation

Radius of the wire
`=0.02" in."=0.02xx2.54xx10^(-2)m=5.08xx10^(-4)m`
Area of cross section of the wire,
`A_(1)=pir^(2)=pi(5.08xx10^(-4))^(2)=81xx10^(-8)m^(2)`
Resistance of the wire,
`R_(0)=(rhol)/(A_(1))=(1.7xx10^(-8)xx0.5)/(81xx10^(-8))Omega` [ = length of the wire = 0.5 m]
`=1.05xx10^(-2)Omega`
Radius of the loop,
`R=(1)/(2pi)=(0.5)/(2pi)=7.96xx10^(-2)m`
Area of the loop,
`A=piR^(2)=3.14xx(7.96xx10^(-2))^(2)`
`=0.02m^(2)`
Here `(dB)/(dt)=100G//s=10^(-2)Wb.m^(-2).s^(-1)`
`therefore` Induced emf,
`e=A.(dB)/(dt)=(0.02)xx10^(-2)=2xx10^(-4)V`
`therefore` From Joule's law, rate of Joule heating,
`H=I^(2)R_(0)=(e^(2))/(R_(0))=((2xx10^(-4))^(2))/(1.05xx10^(-2))=3.8xx10^(-6)W`