SATHEE: Electro Magnetic Induction And Alternating Currents Question 524

Electro Magnetic Induction And Alternating Currents Question 524

Question: A solenoid has 2000 turns wound over a length of 0.30 metre. The area of its cross-section is $1.2 \times 10^{- 3} m^{2}$ . Around its central section, a coil of 300 turns is wound. If an initial current of 2 A in the solenoid is reversed in 0.25 sec, then the e.m.f. induced in the coil is [NCERT 1982; MP PMT 2003]

Options:

A) $6 \times 10^{- 4} V$

B) $4.8 \times 10^{- 3} V$

C) $6 \times 10^{- 2} V$

D) 48 mV

Show Answer

Answer:

Correct Answer: D

Solution:

Induced emf $e = M \frac{d i}{d t} = \frac{μ_{0} N_{1} N_{2} A}{l} . \frac{d i}{d t}$

$= \frac{4 π \times 10^{- 7} \times 2000 \times 300 \times 1.2 \times 10^{- 3}}{0.30} \times \frac{| 2 - (- 2) |}{0.25}$ $= 48.2 \times 10^{- 3} V = 48 m V$

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Electro Magnetic Induction And Alternating Currents Question 524

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