Respuesta :
Answer:
90 V
Explanation:
Parameters given:
Number of turns, N = 15 turns
Area of coil, A = [tex]0.4 m^2[/tex]
Magnetic field, B = 0.75 T
Time taken, t = 0.05 s
The average induced EMF in a coil due to the presence of a magnetic field is given as:
V = [tex]\frac{-NBA}{t}[/tex]
[tex]V = \frac{-15 * 0.75 * 0.4}{0.05}\\ \\\\V = -90 V[/tex]
The magnitude of the average induced EMF will be:
[tex]|V| = |-90| V = 90 V[/tex]
The magnitude of the average induced emf in the coil is 90V.
Calculation of the magnitude:
Here magnitude should be either distance or quantity. It represents the direction or size where an object should moves in the sense of motion
Since
Number of turns, N = 15 turns
Area of the coil, A = 0.4m^2
Magnetic field, B = 0.75 T
Time taken, t = 0.05 s
Now the magnitude is
= -NBA /t
= -15*0.75*0.4/0.05
= -90V
= 90V
Therefore, we can conclude that The magnitude of the average induced emf in the coil is 90V.
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