An air-cored solenoid with length 30 cm,

Solution:

Length of the solenoid, l = 30 cm = 0.3 m

Area of cross-section, A = 25 cm2 = 25 × 10−4 m2

Number of turns on the solenoid, N = 500

Current in the solenoid, I = 2.5 A

Average back emf, $e=\frac{d \phi}{d t}$    ...(1)

Where,

$d \phi=$ Change in flux

= NAB … (2)

Where,

B = Magnetic field strength

$=\mu_{0} \frac{N I}{l}$   ..(3)

Where,

$\mu_{0}=$ Permeability of free space $=4 \pi \times 10^{-7} \mathrm{~T} \mathrm{~m} \mathrm{~A}^{-1}$

Using equations (2) and (3) in equation (1), we get

$e=\frac{\mu_{0} N^{2} I A}{l t}$

$=\frac{4 \pi \times 10^{-7} \times(500)^{2} \times 2.5 \times 25 \times 10^{-4}}{0.3 \times 10^{-3}}=6.5 \mathrm{~V}$

Hence, the average back emf induced in the solenoid is 6.5 V.