First slide

Inductance

Question

A long solenoid of length L, cross-section A having N₁ tums has wound about its centre is small coil of N₂ turns as shown in fig.(10). Then the mutual inductance of two circuits is

Moderate

A

$\frac{μ_{0} A (N_{1} / N_{2})}{L}$
B

$\frac{μ_{0} A (N_{1} N_{2})}{L}$
C

$μ_{0} {AN}_{1} N_{2} L$
D

$\frac{μ_{0} {AN}_{1}^{2} N_{2}}{L}$

Solution

Magnetic flux at the centre of solenoid
$B_{1} = μ_{0} (N_{1} / L) i_{1}$
Magnetic flux through each turn of the coil of area A
$Φ_{1} = B_{1} A = \frac{μ_{0} N_{1} i_{1}}{L} \times A$
Magnetic flux linked with the coil of turns N₂
$Φ_{2} = Φ_{1} \times N_{2} = \frac{μ_{0} N_{1} N_{2} i_{1} A}{L}$
According to the definition of mutual inductance
$\begin{array}{l} Φ_{2} = {Mi}_{1} \\ ∴ {Mi}_{1} = \frac{μ_{0} N_{1} N_{2} i_{1}}{L} A \\ or M = \frac{μ_{0} N_{1} N_{2} A}{L} \end{array}$

Get Instant Solutions

When in doubt download our app. Now available Google Play Store- Doubts App

Recieve an sms with download link

Doubts App

OR

SCAN ME

Doubts App

Doubts App