Question

A solenoid of length 50cm with 20 turns per centimeter and area of cross section $40\;{\text{c}}{{\text{m}}^2}$square completely surrounds another coaxial solenoid of the same length, area of the cross section $25\;{\text{c}}{{\text{m}}^{\text{2}}}$ with $25$ turns per centimeter calculate the mutual inductance of the system.
(A) $9.78\;{\text{mH}}$
(B) $7.85\;{\text{mH}}$.
(C) $8.90\;{\text{mH}}$.
(D) $6.8\;{\text{mH}}$.

Answer 1

In this question, we need to find the expression for the mutual inductance of the system. We know that the mutual induction depends on the number of turns, area of cross-section, and the length of the conductor then substitute all the values in the expression to calculate the mutual inductance of the system.

Complete step by step answer:
We are given a solenoid of length $50\;{\text{cm}}$ and the area of cross section of solenoid is $40\;{\text{c}}{{\text{m}}^2}$ and that is having $20$ turns per centimeter. An area of similar solenoid is $25\;{\text{c}}{{\text{m}}^{\text{2}}}$ and that is having the same length with $25$ turns per centimeter.
Consider all the given values and assume that ${r_1}$ be the radii of ${1^{{\text{st}}}}$ solenoid and ${r_2}$ be the radii of ${2^{{\text{nd}}}}$ solenoid.
And assuming ${n_1}$ be the number of turns of ${1^{{\text{st}}}}$ solenoid and ${n_2}$ be the number of turns of ${2^{{\text{nd}}}}$ solenoid and ${N_{\text{1}}}$ be the total no of turns and ${1^{{\text{st}}}}$ solenoid ${N_{\text{2}}}$ be the total no of turns ${2^{{\text{nd}}}}$ solenoids and each of length $l$.
The mutual inductance between the two solenoids is the interaction of ${1^{{\text{st}}}}$ solenoid magnetic field on another solenoid because of this interaction it induces of unit current flowing in the ${2^{{\text{nd}}}}$ solenoid.
The ${2^{{\text{nd}}}}$solenoids carry current ${I_{\text{2}}}$.So, the current sets up magnetic flux ${\phi _1}$ through the ${1^{{\text{st}}}}$ solenoid.
Write the expression for the total flux.
${N_1}{\phi _1} = {{\text{M}}_{12}}{I_2}$
Where, ${{\text{M}}_{12}}$ is the mutual inductance of the ${1^{{\text{st}}}}$ solenoid with respect to ${2^{{\text{nd}}}}$ solenoid.
The total flux linkages with the ${1^{{\text{st}}}}$ solenoid,
$\Rightarrow {{\text{M}}_{12}} = {\mu _0}{n_1}{n_2}{A_2}l$
$\Rightarrow {{\text{M}}_{12}} = {{\text{M}}_{21}}$
The total flux of the ${1^{{\text{st}}}}$ solenoid is equal to the total flux of ${2^{{\text{nd}}}}$ solenoid.
Converting the values into standard units,
$\Rightarrow {l_1} = {\text{50}}\;{\text{cm}}$
$\Rightarrow {l_1} = {\text{50}}\;{\text{cm}}\left( {\dfrac{{1\;{\text{m}}}}{{{\text{100}}\;{\text{cm}}}}} \right)$
$\Rightarrow {l_1} = {{50 \times 1}}{{\text{0}}^{{\text{ - 2}}}}\;{\text{m}}$
${n_{\text{1}}} = {\text{2000}}\;{\text{turns per meter}}$
$\Rightarrow {A_{\text{1}}} = {{40 \times 1}}{{\text{0}}^{ - {\text{4}}}}\;{{\text{m}}^{\text{2}}}$
${n_{\text{2}}} = {\text{2500}}\;{\text{turns per meter}}$
$\Rightarrow {A_{\text{2}}} = {\text{2}}{{.5 \times 1}}{{\text{0}}^{ - {\text{4}}}}\;{{\text{m}}^{\text{2}}}$

$\Rightarrow {{\text{M}}_{12}} = {\mu _0}{n_1}{n_2}{A_2}l$
Substituting the corresponding values,
$\Rightarrow {{\text{M}}_{{\text{12}}}} = 4\left( {3.14 \times {{10}^{ - 7}}} \right)\left( {2000} \right) \times \left( {2500} \right)\left( {50 \times {{10}^{ - 2}}} \right)\left( {25 \times {{10}^{ - 4}}} \right)$
On simplification,
$\Rightarrow {{\text{M}}_{{\text{12}}}} = \left( {12.56 \times {{10}^{ - 14}}} \right)\left( {2000} \right)\left( {2500} \right)\left( {50} \right)\left( {25 \times {{10}^{ - 2}}} \right)\left( {{{10}^{ - 4}}} \right)$
On further simplification,
$\Rightarrow {{\text{M}}_{{\text{12}}}} = {\text{7}}{\text{.85}}\;{\text{mH}}$

Therefore, the mutual inductance of the system is ${\text{7}}{\text{.85}}\;{\text{mH}}$. So, option (B) is correct.

Note:
We need to be careful about the units of the given variables, first convert all the values into a standard unit that is from centimeter to meter. Do not substitute values without converting the units into the standard units.