Question

if the rate of change of current of 2A/s induces an emf of 10mv in a solenoid what is the self inductance of the solenoid​

Answer 1

Answer:

very easy 5milli Henry is the answer

Explanation:

Formula is

E=L∆I/∆t

L=E∆t/∆I

L=10m/2

L=5millihenry

Answer 2

The required value of self inductance of the solenoid is $5mH$.

Explanation:

Given,

Rate of change of current, $\frac{dI}{dt} = 2A/s$

Induced emf, $e = 10mV = 10$ × $10^{-3} V$

We know that,

emf, $e = L\frac{dI}{dt}$

∴  $10$ × $10^{-3} = L (2)$

or, $L = \frac{10(10^{-3}) }{2}$

∴  $L= 5 mH$

Hence, the required value of self inductance of the solenoid is $5 mH$.