Question

(a) An MRI technician moves his hand from a region of very low magnetic field strength into an MRI scanner’s 2.00 T field with his fingers pointing in the direction of the field. Find the average emf induced in his wedding ring, given its diameter is 2.20 cm and assuming it takes 0.250 s to move it into the field. (b) Discuss whether this current would significantly change the temperature of the ring.

Short answer

1. The value of electromagnetic force is ( EMF) \({\rm{3}}\;{\rm{mV}}\).
2. The temperature is going to be cool.

Step-by-step solution

Concept Introduction

When no current flows, the electromotive force (EMF) is equal to the terminal potential difference. Although both EMF and terminal potential difference (V) are measured in volts, they are not the same. The quantity of energy (E) provided by the battery to each coulomb of charge (Q) flowing through is referred to as the EMF (\({\rm{\varepsilon }}\)).

Calculate the value of EMF

(a)

Knowing the definition of the flux, the value of the electromotive force by stressing the word value, we indicate that we are ignoring the minus related to the direction of the emf will be provided as

\(\begin{aligned}{}{\rm{\varepsilon }} &= \frac{{{\rm{\Delta \Phi }}}}{{{\rm{\Delta t}}}}\\ &= \frac{{{\rm{BA}}}}{{\rm{t}}}\end{aligned}\)

since the flux changed positively the area's value grew and this was from zero to our supplied value We can alternatively assume that the area of a circular cross-section with a known diameter is equal to,

\(\begin{aligned}{}{\rm{A}} &= {\rm{\pi }}{{\rm{R}}^{\rm{2}}}\\ &= \frac{{{\rm{\pi }}{{\rm{D}}^{\rm{2}}}}}{{\rm{4}}}\end{aligned}\)

We derive the parametrical solution by substituting.

\({\rm{\varepsilon }} = \frac{{{\rm{\pi B}}{{\rm{D}}^{\rm{2}}}}}{{{\rm{4t}}}}\)

In terms of numbers, we'll have

\(\begin{aligned}{}{\rm{\varepsilon }} &= \frac{{{\rm{\pi }} \times \left( {{\rm{2}}{\rm{.00}}\;{\rm{T}}} \right) \times {{\left( {{\rm{2}}{\rm{.20}}\;{\rm{cm}}\left( {\frac{{{{10}^{ - 2}}\;{\rm{m}}}}{{1\;{\rm{cm}}}}} \right)} \right)}^{\rm{2}}}}}{{{\rm{4}} \times \left( {{\rm{0}}{\rm{.250}}\;{\rm{s}}} \right)}}\\ &= {\rm{3}} \times {\rm{1}}{{\rm{0}}^{ - 3}}\;{\rm{V}}\left( {\frac{{1\;{\rm{mV}}}}{{{\rm{1}}{{\rm{0}}^{ - 3}}\;{\rm{V}}}}} \right)\\ &= 3\;{\rm{mV}}\end{aligned}\)

Therefore, the EMF is \(3\;{\rm{mV}}\).

Change in temperature

(b)

While we will not compute this because it is a task in the following exercise, we can state that the temperature increase is most likely extremely little. This is because, while the power used at this entrance may be quite high (up to hundreds of watts), the amount of time this heating power is used is quite short.

As a result, without precision instruments, we're probably looking for hardly perceptible temperature variations.

Therefore, the temperature is going to be cool.