Question

A constant current of I = 15 A exists in a solenoid whose inductance is L = 26.0 H. The current is then reduced to zero in a certain amount of time. (a) If the current goes from 15 to 0 A in a time of 75 ms, what is the emf induced in the solenoid? =

(b) How much electrical energy is stored in the solenoid? (Consider the initial energy stored.) E =

Answer 1

The emf induced in the solenoid when the current is switched off can be calculated using Faraday's law of electromagnetic induction. The energy stored in the solenoid at full current can be calculated using the formula for inductance. The rate at which energy must be dissipated to switch off the current in a certain time can be calculated using the formula for power.

Step-by-step explanation:

(a) The induced emf can be calculated using Faraday's law of electromagnetic induction: emf = -L · dI/dt, where L is the inductance and dI/dt is the rate of change of current. Substituting the given values: emf = -25.0 H · (-100 A/0.0800 s) = 31250 V.

(b) The energy stored in an inductor can be calculated using the formula: E = 1/2 · L · I^2, where L is the inductance and I is the current. Substituting the given values: E = 0.5 · 25.0 H · (100 A)^2 = 125000 J.

(c) The rate at which energy is dissipated can be calculated as P = E/t, where P is power, E is energy, and t is time. Substituting the given values: P = 125000 J/0.0800 s = 1.56 MW.

Shutting down the solenoid quickly is difficult because a large amount of energy needs to be dissipated in a short amount of time, resulting in a high power requirement.