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"The superconducting magnet in a magnetic resonance imaging system consists of a solenoid with 5.6 H of inductance. The solenoid is energized by connecting it to a 1.2 V DC power supply. Starting from zero, how long does it take for the solenoid to reach its operating current of 100 A?"

Respuesta :

Olajidey

Answer:

7.8min

Explanation:

The potential difference across an inductor inductance L is

[tex]v_L = L\frac{\Delta i_L}{\Delta t}[/tex]

[tex]\Delta i_L[/tex] is change in curent in a time [tex]\Delta t[/tex]

so, time taken for the solenoid to reach it operating current is

[tex]\Delta t = \frac{L}{v_L} \Delta i_L[/tex]

Given that,

Inductance of solenoid is L = 5.6H

Voltage across solenoid is v(L) = 1.2L

Δi(L) = 100A

substitute the values in the formula

Time taken for the solenoid to reach it operating current is

[tex]\Delta t = \frac{(5.6H)}{1.2V} (100A)[/tex]

= 466.6s

convert to minute

= 466.6 /60

= 7.77min

≅ 7.8min

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