Question

A solenoid coil with 25 turns of wire is wound tightly around another coil with 300 turns. The inner solenoid is 25.0 cm long and has a diameter of 2.00 cm. At a certain time, the current in the inner solenoid is 0.120 A and is increasing at a rate of$\mathbf{1}\mathbf{.}\mathbf{75}\mathbf{\times }{\mathbf{10}}^{\mathbf{3}}\mathit{A}\mathbf{/}\mathit{s}$. For this time, calculate: (a) the average magnetic flux through each turn of the inner solenoid; (b) the mutual inductance of the two solenoids; (c) the emf induced in the outer solenoid by the changing current in the inner solenoid.

Short answer

(a) The average magnetic flux through each turn of the inner solenoid is$5.68\times {10}^{-8}Wb$.

(b) the mutual inductance of the two solenoids is$1.18\times {10}^{-5}H$.

(c) the emf induced in the outer solenoid by changing current in inner solenoid is$-0.0207V$.

Step-by-step solution

Define magnetic flux and mutual induction

Magnetic flux is defined as the number of magnetic field lines passing through a given closed surface. It provides the measurement of the total magnetic field that passes through a given surface area.

When changing current in one circuit changes the magnetic flux in other circuit, an emf is induced in the latter circuit, this phenomena is called mutual induction.The emf is directly proportional to the time rate of current in former circuit. The constant so introduced removing the proportionality is called mutual inductance.

${\mathrm{\epsilon }}_2=-\mathrm{M}\frac{{\mathrm{dI}}_1}{\mathrm{dt}}$where ${\epsilon }_2$is emf induced in second coil, M is the mutual inductance and $\frac{d{I}_1}{dt}$is the time rate of current in first coil.

Mutual inductance is a measure of the mutual inductionbetween two magnetically linked circuits, given as the ratio of the induced emf to the rate of change of current producing it.

It is expressed as,

$M=\frac{{\varphi }_{B2}}{I_1}{N}_2$ where ${\varphi }_{B2}$is magnetic flux linked with second coil having turns $N_2$when current$I_1$is flowing in first coil.

Mutual inductance between two coils depend on area of cross-section, number of turns in each coil, Space between the two coils, Permeability of medium between the two coils and Length (in case of the solenoid).

calculate

A solenoid coil with $N_2=25$turns of wire is wound tightly around another coil with $N_1=300$turns. The inner solenoid is $L=25$long and has a diameter of $2r=2.00cm$. At a certain time, the current in the inner solenoid is $I_1=0.120A$and is increasing at a rate of

$\frac{{\mathrm{dI}}_1}{\mathrm{dt}}$$=1.75\times {10}^{3}A/s.$

.

(a) For this time, magnetic field at center of solenoid is,

$B_1=\frac{{\mu }_o{N}_{1}I}{L}$

$$\begin{gathered} B_1=\frac{4\pi \times {10}^{-7}\times 300\times 0.120}{0.250} \\ {B}_1=1.81\times {10}^{-4}T \end{gathered}$$

So, the magnetic flux linked through each turn of inner solenoid is,

$$\begin{gathered} {\varphi }_B=B_1\times \pi r^2 \\ {\varphi }_B=1.81\times {10}^{-4}\times \pi {\left(\frac{0.0200}{2^2}\right)}^2 \\ {\varphi }_B=5.68\times {10}^{-8}Wb \end{gathered}$$

(b) mutual inductance of the combination is calculated as,

$M=\frac{{\mu }_0{N}_1{N}_2\pi {r_1}^2}{L}$

$$\begin{gathered} M=\frac{4\pi \times 10-7\times 300\times 25\times \pi (0.02/2{)}^2}{0.25} \\ M=1.18\times {10}^{-5}H \end{gathered}$$

(c) the emf induced in the outer solenoid by the changing current in the inner solenoid is calculated as,

$\begin{array}{l}{\epsilon }_2=-M\frac{d{I}_1}{dt}\\ {\epsilon }_2=-1.18\times {10}^5\times 1750\\ {\epsilon }_2=-0.0207V\end{array}$

Therefore, (a)The average magnetic flux through each turn of the inner solenoid is $5.68\times {10}^{-8}Wb$;(b) the mutual inductance of the two solenoids is $5.68\times {10}^{-8}Wb$;(c) the emf induced in the outer solenoid by changing current in inner solenoid is$-0.0207V$