Question

The current in a coil changes from 1 mA to 5 mA in 4 ms. ? If the coefficient of self-inductance of the 'coil is 10 mH,. the magnitude of "self-induced" emf is?

• A. 10 mV
• B. 5 mV
• C. 2.5 mV
• D. 1 mV

Answer 1

The Correct Option is A

We, know, $\varepsilon = \frac{d \phi }{dt}$ and $\phi = LI$
$\therefore \:\:\varepsilon = - L \frac{dI}{dt} $ or , $|\varepsilon | = L \frac{(I_2 - I_12)}{(t_2 - t_1)}$
Here, $L = 10 mH = 10 \times10^{-3} H = 10^{-2} H $
$I_{2} = 5 mA = 5 \times 10^{-3} A$
$ I_{1} = 1 mA = 1 \times 10^{-3}A$
$ t_{2} -t_{1} = 4ms =4 \times 10^{-3} s$
$ =\frac{10^{-2} \times \left(5 \times 10^{-3} - 1 \times 10^{-3}\right)}{4 \times 10^{-3}}$
$ = 10^{-2} = 10 mV$