First slide

Inductance

Question

A solenoid has 2000 turns wound over a length of 0.30 m. The area of its cross-section is 1 .2 x 10^-3 m². Around its central section of a coil of 300 turns is wound. If an initial current of 2 A in the solenoid is reversed in 0.25 sec., the e.m.f. induced in the coil is equal to

Difficult

A

6 x 10^-4 volt
B

4.8 x 10^-2 volt
C

6 x 10^-2 volt
D

48 kV

Solution

Magnetic flux at the centre of solenoid
$B_{1} = [μ_{0} N_{p} i_{1} / l] . . . . . . (1)$
Magnetic flux linked with the coil of N_s turns wound at the centre of solenoid
$Φ_{2} = N_{S} B_{1} A_{2}$
(A₂ = area of cross-section of the coil)
$= N_{s} (\frac{μ_{0} N_{p} i}{l}) A_{2} . . . . . . . (2)$
According to definition of mutual inductance
$\begin{array}{l} Φ_{2} = {Mi}_{1} \\ ∴ {Mi}_{1} = N_{S} (\frac{μ_{0} N_{p} i}{l}) A_{2} \\ or M = \frac{μ_{0} N_{p} N_{s} A_{2}}{l} \end{array}$
In the given problem
$\begin{array}{l} M = \frac{(4 π \times 10^{- 7}) \times 2000 \times 300 \times (1 \cdot 2 \times 10^{- 3})}{0 \cdot 30} \\ e = M \frac{di}{dt} \\ \frac{(4 π \times 10^{- 7}) \times 2000 \times 300 \times (1 \cdot 2 \times 10^{- 3})}{0 \cdot 30} \times \frac{4}{0 \cdot 25} \\ = 4 \cdot 8 \times 10^{- 2} volt \end{array}$

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