Ncert Exercises & Solutions

Q 15. We have an air-cored solenoid having a length of 30 cm, whose area is $25 {cm}^{2},$ and the number of turns is 500. And the solenoid has carried a current of 2.5 A. Suddenly the current is turned off and the time is taken for it is $10^{- 3} s$ . What would be the average value of the induced back-emf by the ends of the open switch in the circuit? (Neglect the variation in the magnetic fields near the ends of the solenoid.)

Length of the solenoid, I = 30 cm = 0.3 mm
Area of cross-section, $A = 25 c m^{2} = 25 \times 10^{- 4} m^{2}$
Number of turns on the solenoid, N = 500
Current in the solenoid, I = 2.5A
Current flows for time, t = $10^{- 3}$ s
Average back emf, $e = \frac{d \emptyset}{d t}$ ...(1)
Where, d $Φ$ = Change in flux = NAB ...(2)
B= Magnetic field strength $= μ_{0} \frac{NI}{l}$ ...(3)
Where, $μ_{0}$ = Permeability of free space $= 4 π \times 10^{- 7} {TmA}^{- 1}$
Using equations (2) and (3) in equation (1), we get

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

$\begin{array}{l} e = \frac{μ_{0} N^{2} IA}{lt} \\ = \frac{4 π \times 10^{- 7} \times (500)^{2} \times 2 .5 \times 25 \times 10^{- 4}}{0 .3 \times 10^{- 3}} = 6 .5 V \end{array}$

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

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