Respuesta :
Explanation:
It is given that,
Area of the loop, [tex]A=8\times 10^{-2}\ m^2[/tex]
Resistance, [tex]R=110\ \Omega[/tex]
Induced current, I = 0.26 A
We know that emf is given by :
[tex]E=A.\dfrac{dB}{dt}[/tex]
From Ohm's law, E = IR
[tex]IR=A.\dfrac{dB}{dt}[/tex]
[tex]\dfrac{dB}{dt}=\dfrac{IR}{A}[/tex]
[tex]\dfrac{dB}{dt}=\dfrac{0.26\ A\times 110\ \Omega}{8\times 10^{-2}\ m^2}[/tex]
[tex]\dfrac{dB}{dt}=357.5\ T/s[/tex]
So, field is changing at the rate of 357.5 T/s. Hence, this is the required solution.
