Respuesta :
Answer:
Thermal energy, [tex]H=8.57\times 10^{-10}\ J[/tex]
Explanation:
It is given that,
Area of loop antenna, [tex]A=3.04\ cm^2=0.000304\ m^2[/tex]
Resistance, [tex]R=6.66\ \mu \Omega=6.66\times 10^{-6}\ \Omega[/tex]
Magnetic field, [tex]B=18.4\ \mu T=18.4\times 10^{-6}\ T[/tex]
The field magnitude drops to zero in 5.43 ms, [tex]t=5.43\times 10^{-3}\ s[/tex]
Due to change in magnetic field, an EMF will induced in the loop which is given by :
[tex]\epsilon=\dfrac{d(BA)}{dt}[/tex]
Also, [tex]\epsilon=IR[/tex]
[tex]IR=\dfrac{d(BA)}{dt}[/tex]
[tex]I=\dfrac{d(BA)}{Rdt}[/tex]
So, [tex]I=\dfrac{18.4\times 10^{-6}\times 0.000304}{6.66\times 10^{-6}\times 5.43\times 10^{-3}}[/tex]
I = 0.154 A
Thermal energy produced is given by,
[tex]H=I^2Rt[/tex]
[tex]H=(0.154)^2\times 6.66\times 10^{-6}\times 5.43\times 10^{-3}[/tex]
[tex]H=8.57\times 10^{-10}\ J[/tex]
So, the thermal energy produced by changing field is [tex]8.57\times 10^{-10}\ J[/tex]. Hence, this is the required solution.
