Question

While keeping area of cross-section of a solenoid same, the number of turns and length of solenoid one both doubled. The self inductance of the coil will be

• A. Halved
• B. Doubled
• C. $\frac{1}{4}$ times the original value
• D. Unaffected

Answer 1

The correct option is B Doubled

The self inductance of a coil is $L=N\frac{\varphi }{I}$ where $N=$ number of turns on coil, $A=$ cross section area and $I=$ current passes through coil.

Now, $\varphi =BA=\left(\frac{{\mu }_{0}NI}{l}\right)A$ where $B=$ magnetic field inside the solenoid and $l=$ length of the solenoid.

Thus, $L={\mu }_0\frac{N^{2}A}{l}$

So, $L^{\prime }={\mu }_0\frac{(2N{)}^{2}A}{\left(2l\right)}=2{\mu }_0\frac{N^{2}A}{l}=2L$