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A loop of radius r = 3.0 cm is placed parallel to the xy-plane in a uniform magnetic field = 0.75 T . The resistance of the loop is 18 Ω. Starting at t = 0, the magnitude of the field decreases uniformly to zero in 0.15 seconds. What is the magnitude of the electric current produced in the loop during that time?
A) 0.79 mA
B) 3.9 mA
C) 1.7 mA
D) 2.1 mA
E) 0.20 mA

Respuesta :

cjmejiab

Answer:

0.79mA

Explanation:

We know that the emf is given by the equation,

[tex]\epsilon = \frac{BA}{\Delta t}[/tex]

We have a circle form, then

[tex]\epsilon = \frac{\pi r^2 B}{\Delta t}[/tex]

[tex]\epsilon = \frac{\pi (0.03)^2(0.75)}{0.15}[/tex]

[tex]\epsilon = 1.41*10^{-2}V[/tex]

We can know calculate the current of the electric field which is given by,

[tex]I \frac{\epsilon}{R} = \frac{1.41*10^{-2}}{18\Omega} = 7.85*10^{-4} = 0.79mA[/tex]

[tex]I=0.79mA[/tex]

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