Question

When the number of the turns in both the two closely wound circular coils kept co-axial one inside the other, are doubled, their mutual inductance becomes

Assume almost equal cross-sectional area and length.

• A. Four times
• B. Two times
• C. Remain same
• D. Sixteen times

Answer 1

The correct option is A Four times

The magnetic field at any point inside the solenoid of length $l$ with total $N_1$ turns carrying a current $i$ is given by the relation,

$B_1=\frac{{\mu }_0{N}_{1}i}{l}$

The magnetic flux through the secondary coil of total $N_2$ turns each of cross-sectional area $A$ is given as,

${\varphi }_{21}=N_2\left(B_{1}A\right)=N_2\left(\frac{{\mu }_0{N}_{1}i}{l}A\right)=\frac{{\mu }_0{N}_1{N}_{2}iA}{l}$

So, mutual inductance,

$M=\frac{{\varphi }_{21}}{i_1}=\frac{\frac{{\mu }_0{N}_1{N}_{2}iA}{l}}{i}=\frac{{\mu }_0{N}_1{N}_{2}A}{l}$

$\Rightarrow M\propto N_1{N}_2$

So, if the number of turns in both the coils are doubled, the mutual inductance will become four times.

Hence, option $\left(A\right)$ is the correct answer.