An air-cored solenoid with length 30 cm, area of cross-section 25 cm2 and number of turns 800, carries a current of 2.5 A. The current is suddenly switched off in a brief time of 10-3s. Ignoring the variation in magnetic field near the ends of the solenoid, the average back emf induced across the ends of the open switch in the circuit would be

(i) zero

(ii) 3.125 volts

(iii) 6.54 volts

(iv) 16.74 volts

Asked by Abhisek | 1 year ago |  65

Solution :-

Right answer is (iv) 16.74 volts

Explanation:-

Magnetic field inside a solenoid

$$B=\mu_0\dfrac{N}{l}I'$$

Initial flux $$\phi=NBA=N\mu_0\dfrac{N}{l}I\;A$$

$$\mu_0\dfrac{N^2}{l}I\;A$$

$$\dfrac{4\pi\times 10^{-7}\times 800\times 800\times 2.5\times 2.5\times 10^{-4}}{0.30}$$

$$16.74 × 10^{– 3} Wb$$

Final flux $$\phi_2=0$$

Average back emf $$|e|=\dfrac{d\phi}{dt}$$

$$\dfrac{ 16.74 × 10^{– 3} Wb}{10^{-3}}$$ = 16.74 V

Answered by Pragya Singh | 1 year ago

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