Electro Magnetic Induction And Alternating Currents Question 482

Question: A resistance less ring has 2 bulbs A and B rated at 2V, 19 W and 2V, 29 W respectively. The ring encloses an ideal solenoid whose magnetic field is as shown. The radius of solenoid is 1 m and the number of turns/length =1000/m. The current changes at rate of 9 A/sec Find the value of P if power dissipated in bulb B is 180 watt.

Options:

A) 4

B) 6

C) 8

D) 11

Show Answer

Answer:

Correct Answer: A

Solution:

  • Resistance of bulb A=v2P=410=0.4

    Resistance of bulb B=v2P=0.2

    emf=dϕdt=ddt(μ0nI×A)

    =μ0n×A×dIdt=107×4π×1000×π(1)2×9

    v=36×103

    I=vReq=36×1030.6=6×102,A Power dissipated through bulb B=I2R =36×104×0.2=7.2×104,watt



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