Answer to Question #230990 in Electricity and Magnetism for Meek

Question

A changing current induces an E.M.F of 10v across a 0.25H inductor. What is the rate at which the current is changing?

Answer 1

 We define the self-inductance of the circuit as $L=\frac{N\Phi_B}{i}$ and the units for the inductance are $1\,H =1\frac{V\cdot s}{A}$.

From Faraday’s law for a coil with N turns, the self-induced emf is $\epsilon = - N \frac{d \Phi_B}{dt}$. Using the prior relation we have:

$\epsilon = - N \frac{d \Phi_B}{dt}=-L\frac{d i}{dt} \implies |\epsilon|=L|\frac{d i}{dt}| \\ |\frac{d i}{dt}|=\cfrac{|\epsilon|}{L}=\cfrac{|10\,V|}{0.25\,H}=\cfrac{10\,V}{0.25\frac{V\cdot s}{A}} \\ |\frac{d i}{dt}|=40 \frac{A}{s}$

In conclusion, the rate of change for the changing current is 40 A/s.

Reference:

• Sears, F. W., & Zemansky, M. W. (1973). University physics.