Answer:
0.652 mA
Explanation:
According to Faraday's Law :
[tex]Emf = - \frac{d \phi}{dt}[/tex]
[tex]= \frac{- \delta \phi }{\delta \ t}[/tex]
[tex]|E| = \frac{A ( \delta B)}{\delta t}[/tex]
where ;
A = [tex]8.60*10^{-4}[/tex]
[tex]\delta B = 3.00 T - 0.5 T = 2.5 T[/tex]
[tex]|E| = \frac{8.60*10^{-4}*2.5}{1.10}[/tex]
[tex]|E| = 1.96*10^{-3}[/tex]
Induced current I = [tex]\frac{|E|}{R}[/tex]
= [tex]\frac{1.96*10^{-3}}{3.0}[/tex]
= [tex]6.52*10^{-4} A[/tex]
= 0.652 mA
Thus, the induced current in the loop of wire over this time = 0.652 A
The induced current in the loop of wire is : 0.625 mA
Given data :
Area of 5 single loop of copper wire = 8.6 cm² = 8.6 * 10⁻⁴ m²
Resistance ( R ) = 3.00
Magnitude of field = 0.5 T
Time = 1.10 sec
According to Faraday's law [tex]Emf = - \frac{d\beta }{dt}[/tex] = [tex]\frac{-d\beta }{dt}[/tex]
Magnitude of E : | E | = [tex]\frac{A(d\alpha) }{dt}[/tex] ----- ( 1 )
where : A = 8.6 * 10⁻⁴ m², [tex]d\alpha = 3 .00 T - 0.5 T = 2.5 T[/tex]
Input values into equation ( 1 )
| E | = 1.96 * 10⁻³
Final step : Calculate the value of induced current
I ( induced current ) = [tex]\frac{|E|}{R}[/tex]
= ( 1.96 * 10⁻³ / 3 ) = 0.625 mA
Hence we can conclude that The induced current in the loop of wire is : 0.625 mA
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