Q67PE - College Physics (Urone)

Question

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 ?

Step-by-Step Solution

Answer

a. The self-inductance is \(200{\rm{ }}H\).

b. The temperature increases is by 5c

Step 1: Concept Introduction

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.

Step 2: Information Given

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

The specific heat value

Step 3: Calculating Self-Inductance

a)

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

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

We can solve for the inductance using this information.

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

In our situation, we will have a numerical advantage.

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

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

Step 4: Calculating Temperature

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

\(Q = cm\Delta T\)

We can solve for the temperature increase as follows:

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

Knowing that all of the energy is turned to heat and using our previously calculated numerical numbers, we can calculate

5c

Therefore, the required solution is 5c