Question

A large research solenoid has a self-inductance of 25.0 H (a) What induced emf opposes shutting it off when 100 Aof current through it is switched off in 80.0ms? (b) How much energy is stored in the inductor at full current? (c) At what rate in watts must energy be dissipated to switch the current off in 80.0ms? (d) In view of the answer to the last part, is it surprising that shutting it down this quickly is difficult?

Short answer

a. The induced electromotive force is 31.25 kV .

b. The energy stored in the inductor is $1.25\times {10}^{5}J$.

c. The power released will be 1.56 MW .

d. No, it is not difficult to switch it off.

Step-by-step solution

Given Data

The self-inductance of the solenoid is M=25.0 H

The time interval when the current is switched off is $∆t=80.0ms\left(\frac{{10}^{-3}s}{1ms}\right)=0.08s$

The change in the current is,$∆l=100A$

Concept Introduction

An inductor is a passive electrical component that counteracts current fluctuations. Coils and chokes are other names for inductors. An inductor is represented by the letter L in electrical notation.

Finding the induced electromotive force(a)

The induced electromotive force can be expressed as,

$\epsilon =-M\frac{∆l}{∆t}$……………..(1)

Where M is self-inductance and $∆l$is the change in the current in the time interval $∆t$.

Substituting the given data in equation (1), we get,

$$\begin{gathered} \epsilon =-\left(25.0H\left(\frac{100A}{0.08s}\right)\right) \\ =-31250V\left(\frac{1kV}{1000V}\right) \\ =-31.25kV \end{gathered}$$

Therefore, the induced electromotive force is$-31.25kV$

Finding energy stored in the inductor(b)

When current runs through an inductor with inductivity M , the energy stored in it can be determined using the expression,

$E=\frac{M{I}^2}{2}$………………(2)

Substituting the given data in equation (2) gives,

$$\begin{gathered} E=\frac{25.0H\times {\left(100A\right)}^2}{2} \\ =1.25\times {10}^{5}J \end{gathered}$$

Therefore, the stored energy is$1.25\times {10}^{5}J$

Finding the power released(c)

If this energy is consistently released over a set of time, the power released can be expressed as,

$P=\frac{E}{t}$………………(3)

Substitute the given data in equation (3), and we get,

$$\begin{gathered} P=\frac{1.25\times {10}^{5}J}{0.08s} \\ =1.56\times {10}^{6}W\left(\frac{1MW}{{10}^{6}W}\right) \\ =1.56MW \end{gathered}$$

Therefore, the power released is$1.56MW$

Explanation(d)

Given the inductor's enormous power, it's not unexpected that turning it down rapidly is difficult.