Question

A solenoid with 200 turns and a cross-sectional area of \(60 \mathrm{~cm}^{2}\) has a magnetic field of \(0.60 \mathrm{~T}\) along its axis. If the field is confined within the solenoid and changes at a rate of \(0.20 \mathrm{~T} / \mathrm{s}\), the magnitude of the induced potential difference in the solenoid will be a) \(0.0020 \mathrm{~V}\). b) \(0.02 \mathrm{~V}\). c) \(0.001 \mathrm{~V}\). d) \(0.24 \mathrm{~V}\).

Short answer

Answer: The magnitude of the induced potential difference in the solenoid is 0.24 V.

Step-by-step solution

Write down Faraday's law of electromagnetic induction

Faraday's law states that the induced electromotive force (EMF) in any closed circuit is equal to the rate of change of the magnetic flux through the circuit. The formula for Faraday's law is: $$ \varepsilon = -N \frac{\Delta \Phi}{\Delta t} $$ Where \(\varepsilon\) is the induced EMF, \(N\) is the number of turns, \(\Delta \Phi\) is the change in magnetic flux, and \(\Delta t\) is the time in which the change occurs.

Calculate the change in magnetic flux

The magnetic flux, \(\Phi\), through the solenoid is given by the product of the magnetic field, \(B\), and the cross-sectional area, \(A\), of the solenoid. So, the change in magnetic flux can be written as: $$ \Delta \Phi = A \Delta B $$ We are given that \(\Delta B = 0.20 \mathrm{~T} / \mathrm{s}\), and the cross-sectional area \(A = 60 \mathrm{~cm}^{2}\). We need to convert the area to \(\mathrm{m}^{2}\) (1 cm = 0.01 m): $$ A = 60 \mathrm{~cm}^{2} \times (0.01)^2 = 0.006 \mathrm{~m}^{2} $$ Now, we can calculate the change in magnetic flux: $$ \Delta \Phi = 0.006 \mathrm{~m}^{2} \cdot 0.20 \mathrm{~T} = 0.0012 \mathrm{~T} \cdot \mathrm{m}^{2} $$

Calculate the induced EMF

Now, we can plug the values into Faraday's law and solve for the induced EMF: $$ \varepsilon = -N \frac{\Delta \Phi}{\Delta t} = -200 \cdot \frac{0.0012 \mathrm{~T} \cdot \mathrm{m}^{2}}{1 \mathrm{s}} $$ Keep in mind that the minus sign in Faraday's law indicates the direction (polarity) of the induced EMF, but we only need its magnitude. So, we can take the absolute value of the expression: $$ |\varepsilon| = 200 \cdot 0.0012 \mathrm{~T} \cdot \mathrm{m}^{2} \cdot \mathrm{s}^{-1} = 0.24 \mathrm{~V} $$

Choose the correct answer

The magnitude of the induced potential difference in the solenoid is \(0.24 \mathrm{~V}\), which matches answer choice (d). So, the correct answer is: (d) \(0.24 \mathrm{~V}\).