Question

If the number of turns in a solenoid is doubled keeping the other factors constant then what will be the new self inductance of the solenoid?
A. Remain unchanged.
B. Be halved
C. Be doubled
D. Become four times

Answer 1

Hint:The magnetic field inside a solenoid is proportional to both the applied current and the number of turns per unit length. There is no dependence on the diameter of the solenoid, and the field strength doesn't depend on the position inside the solenoid, i.e., the field inside is constant.

Formula used:
\[B = {\mu _0}nI\]
This tells us about the magnetic field inside solenoid. Here B is the magnetic field, n is the number of turns and I is the current flowing.

Complete answer:
As we know that the magnetic field inside solenoid is directly proportional to the number of turns of coil and the current flowing through it.
\[B = {\mu _0}nI\]
So, the flux inside the solenoid \[ = B(nl)A \\ \]
flux inside the solenoid= \[B(nl)\pi {r^2}\\ \]
flux inside the solenoid =\[ {\mu _0}{n^2}Il\pi {r^2}\]
This is the total flux in a solenoid.

As we know that self inductance \[ = \dfrac{\phi }{I}\] .Here \[\phi \] is the flux, I is the current flowing. So, the coefficient of self inductance will be \[{\mu _0}{n^2}l\pi {r^2}\]
As from the above relation, we can say that the coefficient of self inductance is directly proportional to the square of number of turns, l is the length and \[{\mu _0},r\] . If we increase the number of turns then the coefficient of self inductance will also increase.
According to the question, if the number of turns of the coil doubled then the coefficient of self inductance would become four times.

Hence, option D is correct.

Note:Here we have taken that nl is the total number of turns in the solenoid. As the length of the solenoid increases then the coefficient of self inductance will also increase as it is directly proportional to the length.