Question

When the number of turns in a coil is doubled without any change in the radius of the coil, its self inductance becomes:
1. Four times
2. Doubled
3. Halved
4. Unchanged

Answer 1

Correct Answer - Option 1 : Four times

CONCEPT:

Self-Induction

• Whenever the electric current passing through a coil changes, the magnetic flux linked with it will also change.
• As a result of this, in accordance with Faraday’s laws of electromagnetic induction, an emf is induced in the coil which opposes the change that causes it.
• This phenomenon is called ‘self-induction’ and the emf induced is called back emf, current so produced in the coil is called induced current.
• Self-inductance of a coil is given as,

\(⇒ L=\frac{{{\mu }_{o}}\pi{{N}^{2}}r}{2}\)

Where μo = Absolute permeability, N = number of turns, and r = radius of the coil

CALCULATION:

Given N₁ = N, N₂ = 2N, and r₁ = r₂ = r

• We know that the self-inductance of a coil is given as,

\(⇒ L=\frac{{{\mu }_{o}}\pi{{N}^{2}}r}{2}\)     -----(1)

By equation 1,

\(⇒ L_1=\frac{{{\mu }_{o}}\pi{{N_1}^{2}}r_1}{2}\)

\(⇒ L_1=\frac{{{\mu }_{o}}\pi{{N}^{2}}r}{2}\)     -----(2)

When the number of turns is in the coil is doubled,

\(⇒ L_2=\frac{{{\mu }_{o}}\pi{{N_2}^{2}}r_2}{2}\)

\(⇒ L_2=\frac{{{\mu }_{o}}\pi{{(2N)}^{2}}r}{2}\)

\(⇒ L_2=\frac{{{4\mu }_{o}}\pi{{N}^{2}}r}{2}\)     -----(3)

By equation 2 and equation 3,

⇒ L₂ = 4L₁

• Hence, option 1 is correct.