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Magnetic resonance imaging instruments use very large magnets that consist of many turns of superconducting wire. A typical such magnet has an inductance of 40 H. When the magnet is initially powered up, the current through it must be increased slowly so as not to "quench" the wires out of their superconducting state. One such magnet is specified to have its current increased from 0 A to 150 A over 180 min. Part A What constant voltage needs to be applied to yield this rate?

Respuesta :

Answer:

V = 0.556V

Explanation:

In magnetic induction

Emf = [tex]\frac{L dI}{dt}[/tex]

This emf is equivalent to voltage applied:

[tex]V = \frac{L dI}{dt}[/tex]

Change in current is 150 - 0 = 150A

change in time is 180min = 180 * 60 = 10800seconds.

Thus,

[tex]V = \frac{40 * 150}{10800} \\V = \frac{6000}{10800}\\ V = 0.556V[/tex]

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