6.15 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.

6.15 Length of the solenoid, l = 30 cm = 0.3 mArea of cross-section, A = 25 cm2 = 25 ×10-4 m2Number of turns on the solenoid, N = 500Current in the solenoid, I = 2.5 ACurrent flowing time, t = 10-3 sWe know average back emf, e = d∅dt …………………. (1)Where d∅= Change in flux = NAB ……………… .(2)B = Magnetic field strength = μ0NIl ………………. (3)Where μ0 = Permeability of free space = 4 π×10-7 T m A-1From equation (1) and (2), we gete = NABdt = NAdt×μ0NIl = μ0N2AIlt = 4π×10-7×5002×25×10-4×2.50.3×10-3 = 6.544 VHence the average back emf induced in the solenoid is 6.5 V

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6.15 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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Payal Gupta | Contributor-Level 10

5 months ago

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

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 = 2.5 A

Current flowing time, t = $10^{- 3}$ s

We know average back emf, e = $\frac{d \emptyset}{d t}$ …………………. (1)

Where $d \emptyset =$ Change in flux = NAB ……………… .(2)

B = Magnetic field strength = $μ_{0} \frac{N I}{l}$ ………………. (3)

Where $μ_{0}$ = Permeability of free space = 4 $π \times 10^{- 7}$ T m $A^{- 1}$

From equation (1) and (2), we get

e = $\frac{N A B}{d t}$ = $\frac{N A}{d t} \times μ_{0} \frac{N I}{l}$ = $\frac{μ_{0} N^{2} A I}{l t}$ &nb

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

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 = 2.5 A

Current flowing time, t = $10^{- 3}$ s

We know average back emf, e = $\frac{d \emptyset}{d t}$ …………………. (1)

Where $d \emptyset =$ Change in flux = NAB ……………… .(2)

B = Magnetic field strength = $μ_{0} \frac{N I}{l}$ ………………. (3)

Where $μ_{0}$ = Permeability of free space = 4 $π \times 10^{- 7}$ T m $A^{- 1}$

From equation (1) and (2), we get

e = $\frac{N A B}{d t}$ = $\frac{N A}{d t} \times μ_{0} \frac{N I}{l}$ = $\frac{μ_{0} N^{2} A I}{l t}$ = $\frac{4 π \times 10^{- 7} \times 500^{2} \times 25 \times 10^{- 4} \times 2.5}{0.3 \times 10^{- 3}}$ = 6.544 V

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

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