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

Hint: 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.