Question

When a magnet in an MRI is abruptly shut down, the magnet is said to be quenched. Quenching can occur in as little as \(20.0 \mathrm{~s}\). Suppose a magnet with an initial field of \(1.20 \mathrm{~T}\) is quenched in \(20.0 \mathrm{~s},\) and the final field is approximately zero. Under these conditions, what is the average induced potential difference around a conducting loop of radius \(1.00 \mathrm{~cm}\) (about the size of a wedding ring) oriented perpendicular to the field?

Short answer

Answer: The average induced potential difference around the conducting loop is approximately \(1.89 \times 10^{-5}\) volts.

Step-by-step solution

Define the given variables and constants

We are given: Initial magnetic field, \(B_1 = 1.20\) T Final magnetic field, \(B_2 \approx 0\) T Time to quench the magnet, \(t = 20.0\) s Radius of the conducting loop, \(r = 1.00\) cm \(= 0.01\) m

Define the formula for Faraday's Law

Faraday's Law states that the induced electromotive force (EMF) is equal to the rate of change of magnetic flux through a closed loop. $$ \varepsilon = - \frac{\Delta \Phi}{\Delta t} $$

Calculate the change in magnetic flux

The magnetic flux \(\Phi\) through a loop of area A perpendicular to the magnetic field is given by: $$ \Phi = B\cdot A $$ Since the loop is a circle, its area can be calculated using $$ A = \pi r^2 $$ The change in magnetic flux \(\Delta \Phi\) is given by the difference between the initial and final fluxes: $$ \Delta \Phi = \Phi_2 - \Phi_1 = B_2A - B_1A $$ As \(B_2 \approx 0\), $$ \Delta \Phi = -B_1A = -1.20 T \cdot \pi (0.01 m)^2 $$

Calculate the average induced potential difference

Now, we can calculate the average potential difference (induced EMF) using Faraday's Law: $$ \varepsilon = - \frac{\Delta \Phi}{\Delta t} = \frac{[1.20 \mathrm{T} \cdot \pi (0.01 \mathrm{m})^2]}{20.0 \mathrm{s}} $$ Calculating the value, $$ \varepsilon \approx 1.89 \times 10^{-5} V $$ So, the average induced potential difference around the conducting loop is approximately \(1.89 \times 10^{-5}\) volts.