Question

One solenoid is centered inside another. The outer one has a length of 50.0 cm and contains 6750 coils, while the coaxial inner solenoid is 3.0 cm long and 0.120 cm in diameter and contains 15 coils. The current in the outer solenoid is changing at 49.2 A>s. (a) What is the mutual inductance of these solenoids

Answer 1

Answer: The mutual inductance of these solenoids is $2.88 \times 10^{-7} H$.

Explanation:

Given: Length = 50.0 cm (1 cm = 0.01 m) = 0.50 m

$N_{1}$ = 6750

$N_{2}$ = 15

Radius = $\frac{0.120 cm}{2} = 0.6 cm = 6 \times 10^{-4} m$

As inner of a solenoid resembles the shape of a circle. So, its area is calculated as follows.

$Area = \pi \times r^{2} = \pi \times (6 \times 10^{-4})^{2}$

Formula used to calculate mutual conductance of two solenoids is as follows.

$M = \frac{\mu_{o} \times A \times N_{1} \times N_{2}}{l}$

where,

M = mutual conductance

A = area

$\mu_{o}$ = relative permeability = $4 \pi \times 10^{-7} Tm/A$

$N_{1}$ = no. of coils in outer solenoid

$N_{2}$ = no. of coils in inner solenoid

l = length

Substitute the values into above formula as follows.

$M = \frac{\mu_{o} \times A \times N_{1} \times N_{2}}{l}\\= \frac{4 \pi \times 10^{-7} Tm/A \times \pi (6 \times 10^{-4})^{2} \times 6750 \times 15}{0.5 m}\\= 2.88 \times 10^{-7} H$

Thus, we can conclude that the mutual inductance of these solenoids is $2.88 \times 10^{-7} H$.