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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 10−3s. Average back emf induced across the ends of the open switch in the circuit is
- a)6.5 V .
- b)4.6 V
- c)4.5 V
- d)5 V3.
Correct answer is option 'A'. Can you explain this answer?
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An air-cored solenoid with length 30 cm, area of cross-section 25cm2an...
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An air-cored solenoid with length 30 cm, area of cross-section 25cm2an...
-4 seconds. Find the average induced emf in the solenoid during this time.
We can use Faraday's law of induction to find the average induced emf in the solenoid:
emf = -N (ΔΦ/Δt)
where N is the number of turns, ΔΦ is the change in magnetic flux, and Δt is the time interval.
Since the current is suddenly switched off, the magnetic flux through the solenoid changes from its maximum value to zero. The maximum value of magnetic flux through the solenoid is given by:
Φ = B A
where B is the magnetic field strength and A is the area of cross-section. Since the solenoid is air-cored, the magnetic field is given by:
B = μ0 N I / L
where μ0 is the permeability of free space, I is the current, and L is the length of the solenoid. Substituting the given values, we get:
B = (4π × 10^-7) × 500 × 2.5 / 0.3 = 1.05 T
Therefore, the maximum magnetic flux through the solenoid is:
Φ = 1.05 × 25 × 10^-4 = 2.625 × 10^-5 Wb
During the brief time of 10^-4 seconds, the change in magnetic flux is equal to the maximum flux, since the current is switched off suddenly. Therefore, ΔΦ = 2.625 × 10^-5 Wb.
Substituting the given values in the formula for emf, we get:
emf = -500 (2.625 × 10^-5) / (10^-4) = -1.3125 V
Therefore, the average induced emf in the solenoid during the brief time of 10^-4 seconds is 1.3125 V. Note that the negative sign indicates that the induced emf opposes the change in current.
We can use Faraday's law of induction to find the average induced emf in the solenoid:
emf = -N (ΔΦ/Δt)
where N is the number of turns, ΔΦ is the change in magnetic flux, and Δt is the time interval.
Since the current is suddenly switched off, the magnetic flux through the solenoid changes from its maximum value to zero. The maximum value of magnetic flux through the solenoid is given by:
Φ = B A
where B is the magnetic field strength and A is the area of cross-section. Since the solenoid is air-cored, the magnetic field is given by:
B = μ0 N I / L
where μ0 is the permeability of free space, I is the current, and L is the length of the solenoid. Substituting the given values, we get:
B = (4π × 10^-7) × 500 × 2.5 / 0.3 = 1.05 T
Therefore, the maximum magnetic flux through the solenoid is:
Φ = 1.05 × 25 × 10^-4 = 2.625 × 10^-5 Wb
During the brief time of 10^-4 seconds, the change in magnetic flux is equal to the maximum flux, since the current is switched off suddenly. Therefore, ΔΦ = 2.625 × 10^-5 Wb.
Substituting the given values in the formula for emf, we get:
emf = -500 (2.625 × 10^-5) / (10^-4) = -1.3125 V
Therefore, the average induced emf in the solenoid during the brief time of 10^-4 seconds is 1.3125 V. Note that the negative sign indicates that the induced emf opposes the change in current.
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An air-cored solenoid with length 30 cm, area of cross-section 25cm2and number of turns 500, carries a current of 2.5 A. The current is suddenly switched off in a brief time of10−3s. Average back emf induced across the ends of the open switch in the circuit isa)6.5 V .b)4.6 Vc)4.5 Vd)5 V3.Correct answer is option 'A'. Can you explain this answer?
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