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An Air-cored Solenoid with Length 30 cm, Area of Cross-section 25 cm^2 And Number of Turns 500, Carries a Current of 2.5 A. the Current is Suddenly Switched off in a Brief Time of 10^−3 s

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प्रश्न

An air-cored solenoid with length 30 cm, area of cross-section 25 cm2 and number of turns 500, carries a current of 2.5 A. The current is suddenly switched off in a brief time of 10−3 s. How much is the average back emf induced across the ends of the open switch in the circuit? Ignore the variation in magnetic field near the ends of the solenoid.

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उत्तर

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

Current flows for time, t = 10−3 s

Average back emf, `"e" = ("d"phi)/("dt")` .........(1)

Where,

`"d"phi` = Change in flux = NAB …........(2)

Where,

B = Magnetic field strength = `mu_0("NI")/"l"` ......(3)

Where,

`mu_0` = Permeability of free space = 4π × 10−7 T m A−1

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

e = `(mu_0"N"^2"IA")/("lt")`

= `(4pi xx 10^-7 xx (500)^2 xx 2.5 xx 25 xx 10^-4)/(0.3 xx 10^-3)`

= 6.5 V

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

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अध्याय 6: Electromagnetic Induction - Exercise [पृष्ठ २३२]

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एनसीईआरटी Physics Part I and II [English] Class 12
अध्याय 6 Electromagnetic Induction
Exercise | Q 6.15 | पृष्ठ २३२

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