Answer:
B) It is doubled.
Explanation:
To determine how the induced electromotive force (emf) in a coil of wire changes when the magnetic field and the number of turns are doubled, and the area is halved, we need to analyze the formula for induced emf according to Faraday's law of electromagnetic induction.
Faraday's law of electromagnetic induction states that the induced emf (ℰ) in a coil is given by:
[tex]\boxed{ \begin{array}{l}\text{\underline{Induced EMF in a Coil:}} \\\\\mathcal{E} = -N \dfrac{d\Phi_B}{dt} \\\\\text{Where:} \\\bullet \ \mathcal{E} \ \text{is the induced electromotive force (emf),} \\\bullet \ N \ \text{is the number of turns in the coil,} \\\bullet \ \Phi_B \ \text{is the magnetic flux,} \\\bullet \ \frac{d\Phi_B}{dt} \ \text{is the rate of change of magnetic flux.}\end{array} }[/tex]
The magnetic flux (Φ) is given by:
[tex]\boxed{ \begin{array}{l}\text{\underline{Magnetic Flux Formula:}} \\\\\Phi = B \cdot A \cdot \cos(\theta) \\\\\text{Where:} \\\bullet \ \Phi \ \text{is the magnetic flux,} \\\bullet \ B \ \text{is the magnitude of the magnetic field,} \\\bullet \ A \ \text{is the area through which the field passes,} \\\bullet \ \theta \ \text{is the angle between the magnetic field and the normal to the surface.}\end{array} }[/tex]
Given the changes:
Let's analyze the new parameters:
The new magnetic flux will be:
[tex]\Longrightarrow \Phi' = B' \cdot A' \cdot \cos(\theta)\\\\\\\\\Longrightarrow \Phi' = 2B \cdot \dfrac{A}{2} \cdot \cos(\theta)\\\\\\\\\therefore \Phi' =B \cdot A \cdot \cos(\theta)[/tex]
So, the magnetic flux 'Φ' remains the same. The induced emf, however, depends on the number of turns and the rate of change of magnetic flux. Since 'Φ' remains the same and 'N' is doubled:
[tex]\Longrightarrow \mathcal{E}' = -N' \dfrac{d\Phi_B'}{dt}\\\\\\\\\Longrightarrow \mathcal{E}' = -2N \dfrac{d\Phi_B}{dt}=\boxed{2\mathcal{E}}[/tex]
The induced emf in the coil is doubled. Thus, the correct answer is:
B) It is doubled.