Correct Answer - Option 4 : 20 mH
Concept:
The formula for inductance of a coil is
\(L = \frac{{\mu {N^2}A}}{l}\)
N is no. of turns
A is the cross-sectional area
l length of the solenoid
\(L \propto \frac{{{N^2}}}{l}\)
\(\frac{{{L_2}}}{{{L_1}}} = {\left( {\frac{{{N_2}}}{{{N_1}}}} \right)^2}\frac{{{l_1}}}{{{l_2}}}\)
Calculation:
Given:
L1 = 5 mH, N1 = 50 turns
Now, The number of turns is doubled i.e.
N2 = 2N1 = 100
l2 = l1 = l
\(\frac{{{L_2}}}{{5 \ mH}} = {\left( {\frac{100}{50}} \right)^2} \)
L2 = 20 mH
