Question

Two toroidal solenoids are wound around the same form so that the magnetic field of one passes through the turns of the other. Solenoid 1 has 700 turns, and solenoid 2 has 400 turns. When the current in solenoid 1 is 6.52 A, the average flux through each turn of solenoid 2 is 0.0320 Wb. (a) What is the mutual inductance of the pair of solenoids? (b) When the current in solenoid 2 is 2.54 A, what is the average flux through each turn of solenoid 1?

Short answer

(a)The mutual inductance of the pair of solenoids is$1.96H$

(b) When the current in solenoid 2 is 2.54 A, the average flux through each turn of solenoid 1 is

Step-by-step solution

Definition of magnetic flux and mutual inductance

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.

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 $I_2$is flowing in first coil.

Calculate magnetic flux and mutual inductance

Two toroidal solenoids are wound around the same form so that the magnetic field of one passes through the turns of the other. Solenoid 1 has $N_1=700$turns, and solenoid 2 has $N_2=400$turns. When the current in solenoid 1 is $I_1=6.52A$, the average flux through each turn of solenoid 2 is ${\varphi }_{B_2}=0.0320Wb$.

(a)The mutual inductance of the pair of solenoids is calculated as,

role="math" localid="1664167297382" $M=\frac{N_2{\varphi }_{B2}}{I_1}$

role="math" localid="1664167303363" $$\begin{gathered} M=\frac{400x0.0320}{6.52} \\ M=1.96H \end{gathered}$$

(b) When the current in solenoid 2 is $I_2=2.54A$, the average flux through each turn of solenoid 1 is,

role="math" localid="1664167581172" $$\begin{gathered} {\varphi }_{B_1}=\frac{M{I}_2}{N_1} \\ {\varphi }_{B_1}=\frac{1.96\times 2.54}{700} \\ {\varphi }_{B_1}=7.11\times {10}^{-3}Wb \end{gathered}$$

Therefore,

(a) The mutual inductance of the pair of solenoids is $1.96H$;

(b) When the current in solenoid 2 is 2.54 A, the average flux through each turn of solenoid 1 is $7.11\times {10}^{-3}Wb$.