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Physics Question on Inductance

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

10 mV

Explanation

Solution

We, know, ε=dϕdt\varepsilon = \frac{d \phi }{dt} and ϕ=LI\phi = LI
∴  ε=−LdIdt\therefore \:\:\varepsilon = - L \frac{dI}{dt} or , ∣ε∣=L(I2−I12)(t2−t1)|\varepsilon | = L \frac{(I_2 - I_12)}{(t_2 - t_1)}
Here, L=10mH=10×10−3H=10−2HL = 10 mH = 10 \times10^{-3} H = 10^{-2} H
I2=5mA=5×10−3AI_{2} = 5 mA = 5 \times 10^{-3} A
I1=1mA=1×10−3AI_{1} = 1 mA = 1 \times 10^{-3}A
t2−t1=4ms=4×10−3st_{2} -t_{1} = 4ms =4 \times 10^{-3} s
=10−2×(5×10−3−1×10−3)4×10−3=\frac{10^{-2} \times \left(5 \times 10^{-3} - 1 \times 10^{-3}\right)}{4 \times 10^{-3}}
=10−2=10mV= 10^{-2} = 10 mV