A very large, superconducting solenoid such as one used in MRI scans, stores \(1.00{\rm{ }}MJ\) of energy in its magnetic field when \(100{\rm{ }}A\) flows. (a) Find its self-inductance. (b) If the coils “go normal,” they gain resistance and start to dissipate thermal energy. What temperature increase is produced if all the stored energy goes into heating the \(1000{\rm{ }}kg\) magnet, given its average specific heat is ?

A passive electrical component that dampens current variations is an inductor. Other names for inductors are coils and chokes. In electrical nomenclature, the letter\(L\)stands in for an inductor.

• Energy stored in solenoid:\(1.00{\rm{ }}MJ\)
• Current in solenoid:\(100{\rm{ }}A\)
• Mass of Magnet:\(1000{\rm{ }}kg\)

The specific heat value

a)

When a current flows through an inductor, the energy stored in it is provided by

\(E = \frac{{L{I^2}}}{2}\)

Wecansolvefortheinductanceusingthisinformation.

\(L = \frac{{2E}}{{{I^2}}}\)

Inoursituation,wewillhaveanumericaladvantage.

\(\begin{array}{c}L = \frac{{2 \times 1 \times {{10}^6}}}{{{{100}^4}}}\\ = 200{\rm{ }}H\end{array}\)

Therefore, the required solution is \(200{\rm{ }}H\).

b)

The amount of energy (heat) required to raise the temperature of mass\(m\)of a certain heat\(c\)material by\(\Delta T\)degrees is given by

\(Q = cm\Delta T\)

Wecansolveforthetemperatureincreaseasfollows:

\(\Delta T = \frac{Q}{{cm}}\)

Knowingthatalloftheenergyisturnedtoheatandusingourpreviouslycalculatednumericalnumbers,wecancalculate

5c

Therefore, the required solution is 5c