SATHEE: Electro Magnetic Induction And Alternating Currents Question 490

Electro Magnetic Induction And Alternating Currents Question 490

Question: A current of 1.5 A flows through a solenoid of length 20.0 cm, cross-section $20.0 c m^{2}$ and 400 turns. The current is suddenly switched off in a short time of 1.0 millisecond. Ignoring the variation in the magnetic field the ends, the average back emf induced in the solenoid is:

Options:

A) 0.3 V

B) 9.6 V

C) 30.0 V

D) 3.0 V

Show Answer

Answer:

Correct Answer: A

Solution:

$e_{b a c k} = L \frac{d i}{d t}$

where $L = \frac{μ_{0} N^{2} A}{l}$
$∴ e_{b a c k} = \frac{μ_{0} N^{2} A}{l} [\frac{1.5 - 0}{1 \times 10^{3}}]$

$= \frac{4 π \times 10^{- 7} {(400)}^{2} \times 20 \times 10^{- 4}}{20 \times 10^{- 2}} \times (1.5 \times 10^{3})$ = 0.3 V

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Electro Magnetic Induction And Alternating Currents Question 490

Question: A current of 1.5 A flows through a solenoid of length 20.0 cm, cross-section 20.0cm2 and 400 turns. The current is suddenly switched off in a short time of 1.0 millisecond. Ignoring the variation in the magnetic field the ends, the average back emf induced in the solenoid is:

Options:

Answer:

Solution:

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Question: A current of 1.5 A flows through a solenoid of length 20.0 cm, cross-section $20.0 c m^{2}$ and 400 turns. The current is suddenly switched off in a short time of 1.0 millisecond. Ignoring the variation in the magnetic field the ends, the average back emf induced in the solenoid is: